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) $y=1; y=\frac1x; y=-\frac1x; y=kx$

The assumption of continuity at $ x = 1 $ is not necessary. As @Hyperplane has shown, we know that:

*

*$ f ( 0 ) = 0 $.

*For $ x \ne 0 $ we have $ f ( - x ) = - f ( x ) $ and thus it suffices to find $ f ( x ) $ for $ x > 0 $.

*$ f \big( x ^ 2 \big) = x f ( x ) $.

Now let $ g ( x ) = f ( x ) - a x $ where $ a $ is a constant real number. Using the original functional equation, it's easy to see that $ g $ satisfies the same equation, i.e.
$$ \big( x ^ 2 - y ^ 2 \big) g ( x y ) = x g \big( x ^ 2 y \big) - y g \big( x y ^ 2 \big) \text . \tag 0 \label 0 $$
Thus $ g $ satisfies the three properties above that hold for $ f $. We can also let $ a = f ( 1 ) $ and have the additional property $ g ( 1 ) = 0 $. Now multiplying \eqref{0} by $ x y $ and using the third property, we get
$$ \big( x ^ 2 - y ^ 2 \big) g \big( x ^ 2 y ^ 2 \big) = g \big( x ^ 4 y ^ 2 \big) - g \big( x ^ 2 y ^ 4 \big) $$
and hence for $ x , y > 0 $ we have
$$ ( x - y ) g ( x y ) = g \big( x ^ 2 y \big) - g \big( x y ^ 2 \big) \text . \tag 1 \label 1 $$
Now letting $ y = \frac 1 x $ in \eqref{1} we get
$$ g \bigg( \frac 1 x \bigg) = g ( x ) \text . \tag 2 \label 2 $$
Again, letting $ y = \frac 1 { x ^ 2 } $ in \eqref{1} and using \eqref{2}, we have
$$ \bigg( x - \frac 1 { x ^ 2 } \bigg) g ( x ) = - g \big( x ^ 3 \big) \tag 3 \label 3 $$
and substituting $ \frac 1 x $ for $ x $ in \eqref{3} and using \eqref{2} we also have
$$ \bigg( \frac 1 x - x ^ 2 \bigg) g ( x ) = - g \big( x ^ 3 \big) \text . \tag 4 \label 4 $$
Subtracting \eqref{3} from \eqref{4} yields
$$ \bigg( x + \frac 1 x + 1 \bigg) \bigg( x - \frac 1 x \bigg) g ( x ) = 0 $$
which shows $ g ( x ) = 0 $ for $ 1 \ne x > 0 $. Therefore $ g $ is the constant zero function and thus $ f ( x ) = a x $.
