Let $g:\mathbb{R}^3 \to \mathbb{R}^2$ be given by $g((x,y,z)) = (g_1(x,y,z)-11, g_2(x,y,z)-3)$. The feasible set is described by $g^{-1}(\{0\})$, and you are trying to find $z$ as a function of $x,y$ locally.
Let me abuse notation and write $g(p,z)$ with $p=(x,y)$ to simplify my life.
Then you need to show that ${\partial g(p,z)) \over \partial p}$ is invertible at a solution $(\hat{p}, \hat{z})$, and the implicit function theorem gives some $\pi$ defined locally around $\hat{z}$ such that $g(\pi(z),z) =0$ for
$z$ near $\hat{z}$, and we have
${\partial \pi(\hat{z})) \over \partial z} = - {\partial g(\hat{p},\hat{z})) \over \partial p}^{-1} {\partial g(\hat{p},\hat{z})) \over \partial z}$.
If we let $\phi(z) = f(\pi(z),z)$ (abusing notation again), then at a solution we will have ${\partial \phi(\hat{z})) \over \partial z} = 0$, using the
chain rule this reduces to
${\partial f(\hat{z},\hat{p})) \over \partial z} = {\partial f(\hat{z},\hat{p})) \over \partial p} {\partial g(\hat{p},\hat{z})) \over \partial p}^{-1} {\partial g(\hat{p},\hat{z})) \over \partial z}$.
(As an aside, when used as a method for numerical optimization, this is known as 'reduced gradients'.)
Given:
$$\begin{equation} \ yz = \ln (x+z) \end{equation} \Rightarrow yz-\ln{(x+z)}=0 \Rightarrow F(x,y,z)=0,$$
we find:
$$z_x=\frac{\partial z}{\partial x} = -\frac{\partial F/\partial x}{\partial F/\partial z}=-\frac{F_x}{F_z}=-\frac{-\frac{1}{x+z}}{y-\frac{1}{x+z}}=\frac{1}{xy+yz-1}.$$
Best Answer
The chain rule says: $$ dG = \frac{\partial G}{\partial x} dx + \frac{\partial G}{\partial y}dy. $$ If the point $(x,y)$ moves along a level set of $G$, then we have $dG=0$. Hence $$ \frac{dy}{dx} = \frac{-\partial G/\partial x}{\phantom{-}\partial G/\partial y}. $$ $$ = -\frac{2xy^4 - 12x^3 y}{x^2 4y^3 -3x^4} $$ and then we can cancel an $x$.
Now let's try implicit differentiation: $$ x^2y^4 - 3x^4y = 0. $$ $$ 2x y^4 + x^2 4y^3 \frac{dy}{dx} - 12x^3y - 3x^4\frac{dy}{dx} =0. $$ Push the two terms not involving the derivative to the other side; then pull out the common factor, which is the derivative; then divide both sides by the other factor. We get $$ \frac{dy}{dx} =\frac{12x^3y - 2xy^4}{x^24y^3 - 3x^4} $$ and it's the same thing.