Finding $\frac{\partial z}{\partial x}$ for $x^2z^2\:+\:xy^2\:−\:z^2\:+\:4yz\:−\:5\:=\:0$ How can I find the  $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ for $x^2z^2\:+\:xy^2\:−\:z^2\:+\:4yz\:−\:5\:=\:0$
I asked my teacher, and he said I should use implicit differentiation.
Also another question is this relationship true?
$\frac{\partial g}{\partial x}\:=\:\frac{\partial z}{\partial x}\times \frac{\partial g}{\partial z},\:\frac{\partial \:z}{\partial \:x}\:=\:\frac{\partial \:g}{\partial \:x}\times \:\frac{\partial \:z}{\partial \:x},\:$ Ie the chain rule for $\frac{dy}{dx}$ for now instead of the "d" I replace with "$\partial$".
I guess I am not allowed to this since I used this in my attempt. And I got the wrong answer. So what is the correct relationship?
And How do I implicit differentiate because now I have 3 variables instead of the usual 2 variables. Also instead of taking the "d", I am taking "$\partial$". Any difference in this aspect? Thanks all - Alice
 A: The "straightforward" derivative $\dfrac{df}{dx}$ is used when we are talking about functions of one variable, and partial, $\dfrac{\partial f}{\partial x}$ - when there is a function $f$ of several variables. There are contexts in which both designations are defined and have different meanings, so it's best not to confuse them.
Due to the implicit function theorem
$$
\frac{\partial z}{\partial x}=-\dfrac{\dfrac{\partial F}{\partial x}}{\dfrac{\partial F}{\partial z}}\qquad
\frac{\partial z}{\partial y}=-\dfrac{\dfrac{\partial F}{\partial y}}{\dfrac{\partial F}{\partial z}},
$$
where
$$
F(x,y,z)=x^2z^2+xy^2−z^2+4yz−5.
$$
We have
$$
\frac{\partial F}{\partial x}=2xz^2+y^2
$$
$$
\frac{\partial F}{\partial y}=2xy+4z
$$
$$
\frac{\partial F}{\partial z}=2x^2z-2z+4y.
$$
Thus
$$
\frac{\partial z}{\partial x}=-\dfrac{2xz^2+y^2}{2x^2z-2z+4y}
$$
$$
\frac{\partial z}{\partial y}=-\dfrac{2xy+4z}{2x^2z-2z+4y}
$$
Update
If you cannot use the implicit function theorem, then you can differentiate both sides with respect to $x$ and express the derivative:
$$
\frac{\partial}{\partial x}\left(x^2z^2+xy^2−z^2+4yz−5\right)=0
$$
$$
2xz^2+x^2\cdot 2z\frac{\partial z}{\partial x}+y^2−2z\frac{\partial z}{\partial x}+4y\frac{\partial z}{\partial x}=0
$$
$$
2xz^2+y^2+\left( 2x^2z-2z+4y \right)\frac{\partial z}{\partial x}=0
$$
$$
\frac{\partial z}{\partial x}=-\frac{2xz^2+y^2}{2x^2z-2z+4y}
$$
