Find an equation of the tangent plane to the given function at the given point. $z = \sin(xy)$ at the point $(x,y) = (2,3\pi /4)$
Work I have done so far:
Partial derivatives:
$$f_x(x,y) = \cos(xy)y \implies f_x\left(2,\frac{3\pi}{4}\right) = \frac{3\pi}{4}\cos\frac{6\pi}{4}\\
f_y(x,y) = \cos(xy)x \implies f_y\left(2,\frac{3\pi}{4}\right) = 2\cos\frac{6\pi}{4}\\
z-0 = \frac{3\pi}{4}\cos\frac{6\pi}{4}\left(x-2\right)+2\cos\frac{6\pi}{4}\left(x-\frac{3\pi}{4}\right)$$
answer: link to incorrect answer

On the left-hand side, you should not have $z-0,$ but $z-f(2,3\pi /4).$  Also, $\cos\frac{6\pi}{4} = \cos\frac{3\pi}{2} = 0,$ so you can make your life a lot easier.
