Partial derivative of an implicit function I am trying to solve this question (sorry if the translation is a bit vague):
$Z(x,y)$ is an implicit function of $x$ and $y$ given in the form of 
$$x^3z^2+\frac{2}{9}y^2\sin(z) = xyz$$
in the neighborhood of $x=2, y=\pi, z=\frac{\pi}{6}$. 
Find $Z_x$ and $Z_y$ at $(x,y) = (2,\pi)$
I know how to partially/totally differentiate, and I know how to find the derivative of a single-variable implicit function. How do I combine it to solve this?
I have the solution in front of me but I can't understand it. Thanks!

Note: This answer was posted before the correction.  However, the same idea still works.
I assume the given equation is
$$x^3z^2 + \frac{2}{9}y^2\sin z = \frac{\pi^2}{3}.$$
Let's now regard $z = Z(x,y)$ as a function of $x$ and $y$ and implicitly (partially) differentiate with respect to $x$:
$$\frac{\partial}{\partial x}\left( x^3z^2 \right) + \frac{\partial}{\partial x}\left(\frac{2}{9}y^2\sin z \right) = \frac{\partial}{\partial x}\left(\frac{\pi^2}{3} \right).$$
That is,
$$3x^2 z^2 + x^3 \frac{\partial}{\partial x}(z^2) + \frac{2}{9}y^2\frac{\partial}{\partial x}(\sin z) = 0,$$
so $$3x^2 z^2 + x^3 (2z)\, Z_x + \frac{2}{9}y^2(\cos z)\,Z_x = 0.$$
Rearranging gives
$$Z_x(x,y) = \frac{-3x^2z^2}{x^3(2z) + \frac{2}{9}y^2\cos z}.$$
Now, we know that when $x = 2$ and $y = \pi$, we have $z = \frac{\pi}{6}$.  Plugging these values in, we find that
$$\begin{align}
Z_x(2, \pi) & = \frac{-3(2^2)(\frac{\pi}{6})^2}{2^3(\frac{2\pi}{6}) + \frac{2}{9}\pi^2\cos\frac{\pi}{6}} \\
& = \frac{ \frac{-\pi^2}{3} }{ \frac{8\pi}{3} + \frac{\pi^2\sqrt{3}}{9} }.
\end{align}$$
Simplifying this expression gives something like $$Z_x(2,\pi) = \frac{-3\pi}{24 + \pi\sqrt{3}},$$ though I might have made a calculation error along the way.
The $Z_y$ case is similar and is left to you.
