How to settle this issue? Our solutions are as follow:

$$when: x \neq0,y \neq 0$$

$$\frac{f(xy)}{(xy)^2} = \frac{f(x)}{x^2} + \frac{f(y)}{y^2}$$

$$let: g(x) = \frac{f(x)}{x^2}$$

$$then :g(xy) = g(x) + g(y)$$

Here, we apply a rather undisciplined approach that forcefully let:

$$g(x) = k \cdot ln|x|$$

then we get a piecewise-defined function:
$$f(x) = k\cdot x^2 \cdot ln|x|, x \neq 0$$
$$f(x) = 0, x = 0$$
emmm, we might as well make some necessary verifications:
$$f(x) = k\cdot x^2 \cdot ln|x|$$
$$f(y) = k\cdot y^2 \cdot ln|y|$$
$$left = f(xy) = k\cdot (xy)^2 \cdot ln(|xy|)$$
$$right = y^2f(x)+x^2f(y) = ky^2 x^2ln|x| + kx^2y^2ln|y| = k(xy)^2 (ln|x| + ln|y|)$$
$$left = right$$

so, we've got points to say
probably:
$$f(x)=\begin{cases}k\cdot x^2 \cdot ln|x|, x \neq 0,\\0, x = 0\end{cases}$$

Note 1:

A rigorous proof includes Cauchy methods, which incorporates continuity conditions. We gonna make it another blog.

Note 2:

What about this:
$$g(x+y) = g(x) \cdot g(y)$$

What if we consider homomorphism and isomorphism in Abstract Algebra?

What ?!

$$ \exists X \in R, s.t. g(X) = 0$$

then, for every given x, we have:

$$g(x) = g(X + (x - X)) = g(X) \cdot g(x - X) = 0$$

if,

$$ \forall x \in R, g(x) \neq 0$$

then g(x) is a function, morphing from R to R:

endomorphism!

$$x, y \in R$$

$$g(x), g(y) \in R^*$$

$$ g: hom : (R, +) \rightarrow (R^*, *) $$

then we consider the kernel, image of g:

同构第一定理！