Find the function satisfying the given condition If $f(xy)=e^{xy-x-y}[e^y f(x) +e^x f(y)]$ and $f'(1)=e$. $f'$ denotes the derivative of function $f(x)$. Find $f(x)$.
I could find that $f(0)=0$ and $f(1)=0$ and then found the derivative the got struck.
 A: $\bf{My\; Solution::}$ Given $$\displaystyle f(xy) = e^{xy-x-y}\left\{e^{y}f(x)+e^{x}f(y)\right\}$$
We Can Write it as $$\displaystyle e^{-xy}f(xy) = e^{-x}f(x)+e^{-y}f(y)$$
Now Let $$e^{-x}f(x) = g(x)\;,$$ Then $$e^{-y}f(y) = g(y)$$ and $$e^{-xy}f(xy) = g(xy).$$
So Functional Equation Convert into $$\displaystyle g(xy)=g(x)+g(y)..........(\star)\color{\red}\checkmark$$
Now Using Partial Differentiation, We Get
Differentiate w r to $x\;,$ Treating $y$ as a Const.
$$\displaystyle g'(xy)\cdot y = g'(x).....................(\star\star)\color{\red}\checkmark$$ 
Differentiate w r to $y\;,$ Treating $x$ as a Const.
$$\displaystyle g'(xy)\cdot x = g'(y).....................(\star \star \star)\color{\red}\checkmark$$
Now Divide These two equation, we get
$$\displaystyle \Rightarrow \frac{y}{x} = \frac{g'(x)}{g'(y)}\Rightarrow y\cdot g(y) = x\cdot g(x) = k$$
So  Let $$\displaystyle x\cdot g'(x) = k\Rightarrow g'(x)=\frac{k}{x}\Rightarrow g(x)=k\cdot \ln (x)+\mathcal{C}$$
Now For Calculation of $\mathcal{C}\;,$ Put $x=y=1$ in $(\star)\color{\red}\checkmark $
So we get $g(1)=2g(1)\Rightarrow g(1) = 0$
So Put $x=1$ in $g(x) = k\cdot \ln (x) +\mathcal{C}$
So We Get $\mathcal{C} = 0\;\;,$ So We Get $g(x) = k\cdot \ln(x)$
So $$e^{-x} \cdot f(x) = k\cdot \ln (x)\Rightarrow f(x) = ke^{x}\ln(x)$$
Now Diff. both Side, We Get $$\displaystyle f'(x) = k\cdot \left[\frac{e^x}{x}+\ln(x)\cdot e^{x}\right]$$
Now Put $x = 1$ in above equation  and Given $f'(1) = e$
We Get value of $k=1$
So Finally we get $$f(x) = e^{x}\cdot \ln(x)$$
