Q) Let $f:R^+\rightarrow R$ be differentiable function with $f(1)=3$ and satisfying $$\int_{1}^{xy}f(t)dt=y\int_{1}^{x}f(t)dt+x\int_{1}^{y}f(t)dt; \forall x,y\in R^+$$, then $f(e)=?$
My Attempt:I tried Leibnitz Rule which states that $$\frac{d}{dx}\int_{\beta(x)}^{\alpha(x)}f(t)dt=f(\alpha(x))\frac{d\alpha(x)}{dx}-f(\beta(x))\frac{d\beta(x)}{dx}$$ but nothing useful seem to come out.
So, please help and $\mathcal THANKS\, IN\, ADVANCE!$
NOTE: ANSWER IS '6'
Differentiating with respect to $x$ gives us $$f(xy)y=yf(x)+\int_1^yf(t)dt$$ Now we differentiate again, this time with respect to $y$, to get: $$f(xy)+f'(xy)xy=f(x)+f(y)$$ Now plug in $x=1$: \begin{align*} f(y)+f'(y)y&=3+f(y)\\ \implies f'(y)&=\frac{3}{y}\\ \implies f(y)&=3\log y+C \end{align*} As $f(1)=3$ we see $C=3$ and hence $f(e)=3+3=6$.