$z = \sin(xy)$ 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 vectors $(1,0,f_x(x_0,y_0))$ and $(0,1,f_y(x_0,y_0))$ are tangent to the surface (in the given point ).
Their cross product $(-f_x,-f_y,1)$ will be normal to the surface and thus to tangent plane.
Now build the equation of a plane, passing through the point $(x_0,y_0,z_0)$ and having the cross product as the normal vector.
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.