Implicit derivative of $$$\sqrt{x y} = x^{2} y^{3} + 1$$$ with respect to $$$x$$$

Differentiate separately both sides of the equation (treat $$$y$$$ as a function of $$$x$$$): $$$\frac{d}{dx} \left(\sqrt{x y{\left(x \right)}}\right) = \frac{d}{dx} \left(x^{2} y^{3}{\left(x \right)} + 1\right)$$$.

Differentiate the LHS of the equation.

The function $$$\sqrt{x y{\left(x \right)}}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \sqrt{u}$$$ and $$$g{\left(x \right)} = x y{\left(x \right)}$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\sqrt{x y{\left(x \right)}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right) \frac{d}{dx} \left(x y{\left(x \right)}\right)\right)}$$

Apply the power rule $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ with $$$n = \frac{1}{2}$$$:

$${\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right)\right)} \frac{d}{dx} \left(x y{\left(x \right)}\right) = {\color{red}\left(\frac{1}{2 \sqrt{u}}\right)} \frac{d}{dx} \left(x y{\left(x \right)}\right)$$

Return to the old variable:

$$\frac{\frac{d}{dx} \left(x y{\left(x \right)}\right)}{2 \sqrt{{\color{red}\left(u\right)}}} = \frac{\frac{d}{dx} \left(x y{\left(x \right)}\right)}{2 \sqrt{{\color{red}\left(x y{\left(x \right)}\right)}}}$$

Apply the product rule $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ with $$$f{\left(x \right)} = x$$$ and $$$g{\left(x \right)} = y{\left(x \right)}$$$:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(x y{\left(x \right)}\right)\right)}}{2 \sqrt{x y{\left(x \right)}}} = \frac{{\color{red}\left(\frac{d}{dx} \left(x\right) y{\left(x \right)} + x \frac{d}{dx} \left(y{\left(x \right)}\right)\right)}}{2 \sqrt{x y{\left(x \right)}}}$$

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$\frac{x \frac{d}{dx} \left(y{\left(x \right)}\right) + y{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{2 \sqrt{x y{\left(x \right)}}} = \frac{x \frac{d}{dx} \left(y{\left(x \right)}\right) + y{\left(x \right)} {\color{red}\left(1\right)}}{2 \sqrt{x y{\left(x \right)}}}$$

Thus, $$$\frac{d}{dx} \left(\sqrt{x y{\left(x \right)}}\right) = \frac{x \frac{d}{dx} \left(y{\left(x \right)}\right) + y{\left(x \right)}}{2 \sqrt{x y{\left(x \right)}}}$$$.

Differentiate the RHS of the equation.

The derivative of a sum/difference is the sum/difference of derivatives:

$${\color{red}\left(\frac{d}{dx} \left(x^{2} y^{3}{\left(x \right)} + 1\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{2} y^{3}{\left(x \right)}\right) + \frac{d}{dx} \left(1\right)\right)}$$

The derivative of a constant is $$$0$$$:

$${\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x^{2} y^{3}{\left(x \right)}\right) = {\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2} y^{3}{\left(x \right)}\right)$$

Apply the product rule $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ with $$$f{\left(x \right)} = x^{2}$$$ and $$$g{\left(x \right)} = y^{3}{\left(x \right)}$$$:

$${\color{red}\left(\frac{d}{dx} \left(x^{2} y^{3}{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) y^{3}{\left(x \right)} + x^{2} \frac{d}{dx} \left(y^{3}{\left(x \right)}\right)\right)}$$

The function $$$y^{3}{\left(x \right)}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = u^{3}$$$ and $$$g{\left(x \right)} = y{\left(x \right)}$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$$x^{2} {\color{red}\left(\frac{d}{dx} \left(y^{3}{\left(x \right)}\right)\right)} + y^{3}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right) = x^{2} {\color{red}\left(\frac{d}{du} \left(u^{3}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)} + y^{3}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right)$$

Apply the power rule $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ with $$$n = 3$$$:

$$x^{2} {\color{red}\left(\frac{d}{du} \left(u^{3}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{3}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right) = x^{2} {\color{red}\left(3 u^{2}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{3}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right)$$

Return to the old variable:

$$3 x^{2} {\color{red}\left(u\right)}^{2} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{3}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right) = 3 x^{2} {\color{red}\left(y{\left(x \right)}\right)}^{2} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{3}{\left(x \right)} \frac{d}{dx} \left(x^{2}\right)$$

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 2$$$:

$$3 x^{2} y^{2}{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{3}{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} = 3 x^{2} y^{2}{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + y^{3}{\left(x \right)} {\color{red}\left(2 x\right)}$$

Simplify:

$$3 x^{2} y^{2}{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 x y^{3}{\left(x \right)} = x \left(3 x \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 y{\left(x \right)}\right) y^{2}{\left(x \right)}$$

Thus, $$$\frac{d}{dx} \left(x^{2} y^{3}{\left(x \right)} + 1\right) = x \left(3 x \frac{d}{dx} \left(y{\left(x \right)}\right) + 2 y{\left(x \right)}\right) y^{2}{\left(x \right)}$$$.

Therefore, we have obtained the following linear equation with respect to the derivative: $$$\frac{x \frac{dy}{dx} + y}{2 \sqrt{x y}} = x y^{2} \left(3 x \frac{dy}{dx} + 2 y\right)$$$.

Solving it, we obtain that $$$\frac{dy}{dx} = \frac{- 4 y^{2} \left(x y\right)^{\frac{3}{2}} + y}{6 x y \left(x y\right)^{\frac{3}{2}} - x}$$$.