, 2015

Answer: Either $f(z) = g(z) = \delta_{z,0}$ or $f(z) = bz$ and $g(z) = az^2$.

Putting $xy = w$ in the second equation gives
$$0 = g\left(\frac{1}{y}f(w) - b\frac{w}{y}\right) = g\left(\frac{1}{y}(f(w) - bw)\right)$$
for all $y \neq 0$. Hence, if $f(w) \neq bw$ for some $w$, it must be that $g(z) = 0$ for $z \neq 0$. Since $g$ must assume the value 1 somewhere, $g(0) = 1$.

The function $g$ now being known, the first equation transforms, for $x = 0$ and $x \neq 0$, respectively, into
$$f\left(\frac{1}{y^2}\right) = 0 \quad \text{and} \quad f(-ax^2) = 0.$$
Consequently, $f(z) = 0$ for $z > 0$ or $z < 0$. Again, $f$ must assume the value 1, and so $f = g$.

There remains the case when $f(z) = bz$ for all $z$. Substitute $y = 1$ into the first equation to find $b(g(x) - ax^2) = 0$, so that $g(z) = az^2$ for all $z$.

One easily verifies that these two possibilities satisfy the requirements. $\square$