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.