Implicit differentiation of equation The equation $2x^4z^2+y^4=x^2-4xy^5z$ defines $z$ implicitly as a function of $x$ and $y$. Find $\frac{\partial z}{\partial x}$ at the point $(1, -1, 0)$
The solution was as follows:
So if we derive with respect to $x$ we get $$8x^3z^2+4x^4z\frac{\partial z}{\partial x} = 2x-4y^5z-4xy^5\frac{\partial z}{\partial x}$$
Plugging inn $(1, -1, 0)$ we get $\frac{\partial z}{\partial x} = \frac{1}{2}$.
But when we differentiate with respect to x, on the last expression they used the chain rule on $x$ and $z$ and treated $y$ as a constant. Shouldn't you also use the chain rule on $y$ and get something like $-4xy^5z-4xy^5\frac{\partial z}{\partial x}-20xy^4z\frac{\partial y}{\partial x}$.
I realize now that since $z=0$ in our point, the last expression would actually fall away and we would get the right answer. But the solution doesn't contain the last expression at all, so I'm confused about whether or not I've misunderstood implicit differentiation.
Help would be greatly appreaciated as I have an exam tomorrow... thank you!
 A: To elaborate on my comment, let's remember what everything means. If we are treating $z$ as a function of $x$ and $y$, then if we see something like
$$
\frac{\partial}{\partial x}\Big[ xy + xz \Big]
$$
then this is equal to
$$
\frac{\partial}{\partial x}\Big[ xy \Big] + \frac{\partial}{\partial x}\Big[ xz \Big].
$$
But $y$ does not depend on $x$, so we treat it as a constant and thus
$$
\frac{\partial}{\partial x}\Big[ xy \Big] = y\cdot\frac{\partial}{\partial x}\Big[ x \Big] = y \cdot 1 = y.
$$
On the other hand, since $z$ is a function of $x$, we use the product rule
$$
\frac{\partial}{\partial x}\Big[ xz \Big] = \frac{\partial}{\partial x}\Big[ x \Big]\cdot z + x\cdot\frac{\partial}{\partial x}\Big[ z \Big] = 1\cdot z + x\cdot \frac{\partial z}{\partial x} = z + x\frac{\partial z}{\partial x}.
$$
Thus
$$
\frac{\partial}{\partial x}\Big[ xy + xz \Big] = y + z + x\frac{\partial z}{\partial x}.
$$
There is no need to do the same computations with $y$, since $y$ is not a function of $x$.
