An algebra problem by Kushal Dey

This functional equation closely resembles the linear-based Cauchy Functional Equation. Let us take $f(x)=Ax+B$ with the condition $f(0)=0$ for the case $x=y=z=0$ . Substituting this function in the original functional equation gives:

$A(x^2 + Ayz + By) + B = Ax^2 + Bx + Ayz + Bz \Rightarrow Ax^2 + A^{2}yz + ABy + B = Ax^2 + Bx + Ayz + Bz \Rightarrow A^{2}yz + ABy + B = Bx + Ayz + Bz$

With a little coefficient matching, we obtain the following relationships:

$A^2 = A \Rightarrow A = 0,1$

$(Ay+1)B = (x+z)B \Rightarrow B = 0$ (since $f(0)=0$ )

which gives us the functions $f(x) = x$ or $f(x) = 0$ . Since we're given that $f(1) \neq 0$ , the latter constant function poses a contradiction. Hence $f(x) = x$ is the only satisfying function. Choice A is the correct option since $f$ is an odd function.

We just need to put some special values of x,y,z to solve this problem.
Put x=0,y=z, f(yf(y))=yf(y) ....(1)
Put x=0,y=1, f(f(z))=z * f(1) .....(2)
Put y=z=0, f(x^2)=xf(x)
=> f(f(x²))=f(xf(x)),
Now using (1) in rhs and (2) in lhs we get,
x²f(1)=xf(x) => f(x)=xf(1) ....(3)
Now expand original equation using (3)
x²f(1)+yf(1)f(z)=x² f(1)+y * f(1) * z
On simplifying, f(z)=z.
You may use higher level calculus method which includes taking partial derivatives but I think this is the most simple method to solve this problem