Implicit derivative of $$$x^{2} y^{2} = 2 x + e^{y}$$$ with respect to $$$x$$$

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

Differentiate the LHS of the equation.

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 $$$g{\left(x \right)} = y^{2}{\left(x \right)}$$$:

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

The function $$$y^{2}{\left(x \right)}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = u^{2}$$$ and $$$g{\left(x \right)} = y{\left(x \right)}$$$.

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

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

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

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

Return to the old variable:

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

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

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

Simplify:

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

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

Differentiate the RHS of the equation.

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

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

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

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

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

The derivative of the exponential is $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:

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

Return to the old variable:

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

Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = 2$$$ and $$$f{\left(x \right)} = x$$$:

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

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

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

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

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

Solving it, we obtain that $$$\frac{dy}{dx} = \frac{- 2 x y^{2} + 2}{2 x^{2} y - e^{y}}$$$.