I think to find a tangent plane I have to find a normal vector to the plane, and know a point on the plane. I know a point is $(-2,3\pi/4, \pi/4)$ but I don't know how to find a normal vector.
Also, I am a bit confused - isn't $f(x,y,z)$ function of 3 variables which means the graph is 4D? So is it tangent "hyperplane"? Or is it regular 2D tangent plane?
Thanks for help
I found the answer from here: https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-b-chain-rule-gradient-and-directional-derivatives/session-37-example/MIT18_02SC_we_17_comb.pdf
The "level surface" is $$x^2 \sin(2y) \cos^2 (z) = -2$$
Then put a new variable $w$ so that $$w = x^2 \sin(2y) \cos^2(z)$$
Now have to find gradient on $w$:
$$\nabla w = \left(\frac{\partial w}{\partial x},\frac{\partial w}{\partial y},\frac{\partial w}{\partial z}\right) = \left(2x\sin(2y)\cos^2(z), 2x^2\cos(2y)\cos^2(z), -2x^2\sin(2y)\sin(z)\cos(z)\right)$$
Now to plug in point $P (-2, 3\pi/4, \pi/4) = (x,y,z)$ to this: $$\nabla w | P = \left(-4\sin(3\pi/2)\cos^2(\pi/4), 8\cos(3\pi/2)\cos^2(\pi/4), -8\sin(3\pi/2)\sin(\pi/4)\cos(\pi/4)\right)$$
so it is $$\nabla w | P = \left(2,0,4\right)$$
this means tangent plane is: $$2(x+2)+4(z-\pi/4) = 0$$
which is same as $$2x + 4z = \pi -4 $$