Implicit derivative of $$$- x^{2} + y^{2} - 4 y + 12 = 0$$$ with respect to $$$x$$$

Differentiate separately both sides of the equation (treat $$$y$$$ as a function of $$$x$$$): $$$\frac{d}{dx} \left(- x^{2} + y^{2}{\left(x \right)} - 4 y{\left(x \right)} + 12\right) = \frac{d}{dx} \left(0\right)$$$.

Differentiate the LHS of the equation.

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

$${\color{red}\left(\frac{d}{dx} \left(- x^{2} + y^{2}{\left(x \right)} - 4 y{\left(x \right)} + 12\right)\right)} = {\color{red}\left(- \frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(y^{2}{\left(x \right)}\right) - \frac{d}{dx} \left(4 y{\left(x \right)}\right) + \frac{d}{dx} \left(12\right)\right)}$$

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

$${\color{red}\left(\frac{d}{dx} \left(12\right)\right)} - \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(4 y{\left(x \right)}\right) + \frac{d}{dx} \left(y^{2}{\left(x \right)}\right) = {\color{red}\left(0\right)} - \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(4 y{\left(x \right)}\right) + \frac{d}{dx} \left(y^{2}{\left(x \right)}\right)$$

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

$${\color{red}\left(\frac{d}{dx} \left(y^{2}{\left(x \right)}\right)\right)} - \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(4 y{\left(x \right)}\right) = {\color{red}\left(\frac{d}{du} \left(u^{2}\right) \frac{d}{dx} \left(y{\left(x \right)}\right)\right)} - \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(4 y{\left(x \right)}\right)$$

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

$${\color{red}\left(\frac{d}{du} \left(u^{2}\right)\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(4 y{\left(x \right)}\right) = {\color{red}\left(2 u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(4 y{\left(x \right)}\right)$$

Return to the old variable:

$$2 {\color{red}\left(u\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(4 y{\left(x \right)}\right) = 2 {\color{red}\left(y{\left(x \right)}\right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - \frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(4 y{\left(x \right)}\right)$$

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 $$$c = 4$$$ and $$$f{\left(x \right)} = y{\left(x \right)}$$$:

$$2 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - {\color{red}\left(\frac{d}{dx} \left(4 y{\left(x \right)}\right)\right)} - \frac{d}{dx} \left(x^{2}\right) = 2 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - {\color{red}\left(4 \frac{d}{dx} \left(y{\left(x \right)}\right)\right)} - \frac{d}{dx} \left(x^{2}\right)$$

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

$$2 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} - 4 \frac{d}{dx} \left(y{\left(x \right)}\right) = 2 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - {\color{red}\left(2 x\right)} - 4 \frac{d}{dx} \left(y{\left(x \right)}\right)$$

Thus, $$$\frac{d}{dx} \left(- x^{2} + y^{2}{\left(x \right)} - 4 y{\left(x \right)} + 12\right) = - 2 x + 2 y{\left(x \right)} \frac{d}{dx} \left(y{\left(x \right)}\right) - 4 \frac{d}{dx} \left(y{\left(x \right)}\right)$$$.

Differentiate the RHS of the equation.

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

$${\color{red}\left(\frac{d}{dx} \left(0\right)\right)} = {\color{red}\left(0\right)}$$

Thus, $$$\frac{d}{dx} \left(0\right) = 0$$$.

Therefore, we have obtained the following linear equation with respect to the derivative: $$$- 2 x + 2 y \frac{dy}{dx} - 4 \frac{dy}{dx} = 0$$$.

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