no. of function

set x=y = 1

we have f(1)f(1)f(z) = 12 f(z) - 16 z

f(1)f(1) ≠12 because otherwise z = 0 for all z, contradiction

So now f(z) = 16z12 - f(1)f(1) = kz

So f is linear

continuing from above post ,

f(z) = kz
putting z=1 ,
f(1)=k

but k = 1612-{f(1)}² = f(1)

thus on solving , f(1) = 2,-4

thus f(x) =2x and f(x)=-4x are the solutions.

there are two function which satisfied R:R