Given:
f(x, y, z) = f(x) * f(y) * f(z) for all x, y, z
f(2) = 4
f'(0) = 5
f(0) ≠ 0

To find:
f'(2)

Solution:
To find f'(2), we need to differentiate the given function f(x) with respect to x and then substitute x = 2.

Differentiating f(x) with respect to x, we get:
f'(x, y, z) = f'(x) * f(y) * f(z)

Now, let's differentiate both sides of the given equation with respect to x:
∂/∂x [f(x, y, z)] = ∂/∂x [f(x) * f(y) * f(z)]

Using the product rule of differentiation, we have:
f'(x, y, z) = f'(x) * f(y) * f(z) + f(x) * f'(y) * f(z) + f(x) * f(y) * f'(z)

Now, substitute x = 2 in the above equation:
f'(2, y, z) = f'(2) * f(y) * f(z) + f(2) * f'(y) * f(z) + f(2) * f(y) * f'(z)

Given f(2) = 4, we can substitute this value:
f'(2, y, z) = f'(2) * f(y) * f(z) + 4 * f'(y) * f(z) + 4 * f(y) * f'(z)

Since this equation holds true for all y and z, we can differentiate it with respect to y and z separately to eliminate the variables y and z.

Differentiating the above equation with respect to y, we get:
∂/∂y [f'(2, y, z)] = ∂/∂y [f'(2) * f(y) * f(z) + 4 * f'(y) * f(z) + 4 * f(y) * f'(z)]

Using the product rule, we have:
0 = f'(2) * f'(y) * f(z) + 4 * f''(y) * f(z) + 4 * f(y) * f'(z)

Now, substitute y = 0 in the above equation:
0 = f'(2) * f'(0) * f(z) + 4 * f''(0) * f(z) + 4 * f(0) * f'(z)

Given f'(0) = 5 and f(0) ≠ 0, we have:
0 = f'(2) * 5 * f(z) + 4 * f''(0) * f(z) + 4 * f(0) * f'(z)

Since this equation holds true for all z, we can differentiate it with respect to z to eliminate the variable z.

Differentiating the above equation with respect to z, we get:
0 = f'(2) * 5 * f'(z) + 4 * f''(0) * f'(z) + 4 * f(0) * f''(z)

Now, substitute z = 0 in the