Find $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$ if $z=f(x,y)$ is defined implicitly by the function $$F(x,y,z)=x^{2}z^{2}+y^{2}-z^{2}+6yz-10=0.$$

Solution First we find the partial derivatives of $F.$ $$\begin{equation*} F_{x}=\dfrac{\partial F}{\partial x}=2xz^{2}\qquad F_{y}=\dfrac{ \partial F}{\partial y}=2y+6z\qquad F_{z}=\dfrac{\partial F}{\partial z} =2x^{2}z-2z+6y \end{equation*}$$

Then we use (4). If $F_{z}=2x^{2}z-2z+6y\neq 0$, $$\dfrac{\partial z}{\partial x}=-\dfrac{2xz^{2}}{2x^{2}z-2z+6y}=-\dfrac{ xz^{2}}{x^{2}z-z+3y}$$

and $$\dfrac{\partial z}{\partial y}=-\dfrac{2y+6z}{ 2x^{2}z-2z+6y}=-\dfrac{y+3z}{x^{2}z-z+3y}$$