<s>[INST] Find all functions $f:\mathbb R \rightarrow \mathbb R$, such that: $(x^2 − y^2)\cdot f(xy) = x\cdot f(x^2y) − y\cdot f(xy^2)$ Find all functions $f:\mathbb R \rightarrow \mathbb R$, such that:
$$(x^2 − y^2)\cdot f(xy) = x\cdot f(x^2y) − y\cdot f(xy^2)$$
for all $x,y \in \mathbb R$
My work so far:
1) $f(0)=0$
2) $y=1; y=\frac1x; y=-\frac1x; y=kx$
 [/INST] Since my first answer contained a mistake so I am posting this revised version which only gives a partial result to the question.
Claim: Assume $f\colon\mathbb R\to\mathbb R$ satisfies $(x^2-y^2)f(xy) = xf(x^2y)-yf(xy^2)$ for all real $x,y$, and is continuous in $1$. Then $f(x) = f(1)x$ for all $x$; i.e. $f$ is linear.
Proof: Plugging in $y=-x$ yields
$$ 0 = xf(-x^3) +xf(x^3) \implies f(x) = -f(-x) \qquad\text{for } x\neq 0 $$
Moreover since $f(0)=0$, as seen by plugging in  $(x,y)=(-1,0)$, we conclude that $f$ is an odd function. Now if we plug in $y=1$ instead we find that for all $x\neq 0$:
$$  (x^2 -1)f(x) = xf(x^2) -f(x) \implies x^2 f(x) = xf(x^2) \implies 
  \boxed{\frac{f(x)}{x} = \frac{f(x^2)}{x^2}}\qquad (I) $$
For $x>0$ we can subsitute $x\to\sqrt{x}$ to find $\frac{f(x)}{x} = \frac{f(\sqrt{x})}{\sqrt{x}} = \frac{f(\sqrt{\sqrt{x}})}{\sqrt{\sqrt{x}}} = \ldots = \frac{f(\sqrt[2^k]{x})}{\sqrt[2^k]{x}}$ for all $k\ge0$ by repeatedly applying the equation to itself. Then we can use the well known fact that $\lim\limits_{n\to\infty}\sqrt[n]{x} =1$ for all $x>0$, which together with the continuity of $f$ in $1$ implies that for $x>0$:
$$ f(1) =\lim_{k\to\infty}\frac{f(\sqrt[2^k]{x})}{\sqrt[2^k]{x}} = \lim_{k\to\infty} \frac{f(x)}{x} = \frac{f(x)}{x} $$
Thus by symmetry we have $f(x)=f(1)x$ for all $x$.
Remark: I wasn't able to deduce continuity of $f$ in $1$, but one possible approach might be to set $y = \frac{1}{x}$ and use $xf(x) = f(x^2)$, a variation of $(I)$, to deduce:
\begin{gather}
  (x^2 - \frac{1}{x^2})f(1) = xf(x) - \frac{1}{x}f(\frac{1}{x}) = f(x^2) - f(\frac{1}{x^2})
\\
\implies \boxed{f(1) = \frac{f(x) - f(\frac{1}{x})}{x-\frac{1}{x}}}
\end{gather}
Which holds not only for positive $x$, but by symmetry for all $x\neq 0$, and then somehow use that $x\approx 1 \iff \frac{1}{x} \approx 1$ to show continuity. Alternatively one might try to find a pathological counter-example that  is not continuous, similarly as is the case with Cauchy's functional equation.
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