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A twice differentiable function$f(x)$ satisfies $$f(x^2y)=x^2f(y)+yf(x^2) ,\forall x,y\gt 0$$ and $f'(1)=1$. From this information we have to find the value of $f''(\frac{1}{7})$. It is also to be deduced what $f(x)$ is.

First I plugged in $x=y=1$ and obtained $f(1)=0$

To make the equation a little workable, I just put $x^2=\frac{1}{y}$ and obtained $$\frac{1}{y}f(y) + yf(\frac{1}{y})=f(1)=0$$

Taking the derivative,$$-f(y) + yf'(y) + y^2f(\frac{1}{y})-yf'(\frac{1}{y})=0$$

Again taking the derivative and simplifying $$y^2f''(y)+2y^2f(\frac{1}{y})+ f''(\frac{1}{y}) -2yf'(\frac{1}{y})=0$$

How should I proceed after this? Or is it that this approach is incorrect?

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Hint: substitute $x^2 \to x\,$, then:

$$ f(xy)=xf(y)+yf(x) \;\;\iff\;\; \frac{f(xy)}{xy}=\frac{f(x)}{x}+\frac{f(y)}{y} $$


[ EDIT ]   Extended hint:  let $\,g(x)=\cfrac{f(x)}{x}\,$ which must be continuous on $\mathbb{R}^+\,$ since $f(x)$ is differentiable. Then the above can be written as $\,g(xy)=g(x)+g(y)\,$, which is related to the Cauchy functional equation. Among continuous functions, the only solutions are of the form $\,g(x) = \lambda \ln x\,$ for some constant $\,\lambda \in \mathbb{R}\,$, so in the end $\,f(x) = \lambda x \ln x\,$.

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