Solve the functional equation $f(xy)=e^{xy-x-y} \big( e^yf(x)+e^xf(y) \big)$ If $f: \Bbb R ^+ \to  \Bbb R$ satisfies $$f(xy)=e^{xy-x-y} \big( e^yf(x)+e^xf(y) \big)$$ for all $x, y \ge 0$ and if $f'(1)=e$, determine $f(x)$.
I'm a beginner. Can someone give me some hints for this sum?
I'm getting $f(1)=0$. Is that right?
 A: \begin{align*}
  f(xy)&=e^{xy-x-y} \left( e^yf(x)+e^xf(y) \right) \\
  f(1) &= e^0 f(1) + e^0 f(1) \\
  f(1) &= 0
\end{align*}
If $x>0$, set $y = \frac{x+h}{x}$, then $xy=x+h$.
\begin{align*}
  f(xy)&= f(x+h) \\
  &= e^{x\frac{x+h}{x}-x-\frac{x+h}{x}} \left( e^\frac{x+h}{x} f(x)+e^xf(\frac{x+h}{x}) \right) \\
  &= e^{h-\frac{x+h}{x}} \left( e^\frac{x+h}{x} f(x)+e^xf(\frac{x+h}{x}) \right) \\
&=e^h f(x) + e^{x+h-\frac{x+h}{x}} f(1+\frac{h}{x})
\end{align*}
\begin{align*}
  \frac{f(x+h)-f(x)}{h} &= \frac{(e^h - 1) f(x) + e^{x+h-\frac{x+h}{x}} f(1+\frac{h}{x})}{h} \\
  &= \frac{e^h - 1}{h} f(x) + \frac{e^{x+h-\frac{x+h}{x}}}{x} \frac{f(1+\frac{h}{x}) - f(1)}{\frac{h}{x}}
\end{align*}
Since $f'(1) = e$, take limit on both sides as $h \to 0$.
\begin{align*}
  f'(x) &= f(x) + \frac{e^{x-1}}{x} f'(1) \\
  &= f(x) + \frac{e^x}{x} \\
  &= f(x) + \frac{e^{x}}{x}
\end{align*}
Let $y = f(x)$.
\begin{align*}
  y'-y &= \frac{e^{x}}{x} \\
  (y e^{-x})' &= \frac{1}{x} \\
  y e^{-x} &= \ln x + C \\
  y &= e^{x} \ln x + C e^x
\end{align*}
Since $y = f(1) = 0, C = 0$.  Hence $y = e^{x} \ln x \forall x > 0$.
Alternative method
I followed Sanchayan Dutta's method.

Why can we do a partial differentiation w.r.t. $x$?

We keep $y$ constant, and treat $x$ as the only independent variable.
Looking at the calucations Sanchayan Dutta's answer, I was puzzled and I wondered where the term $e^{xy}$ was.  Why are our calculations so different?  Why can we get the right answer despite different calculations?  As a result, I decided to do it myself and to add this section.
\begin{equation*}
   f(xy)=e^{xy-x-y} \left( e^yf(x)+e^xf(y) \right)
\end{equation*}
Partial differentiation with respect to $x$ gives
\begin{align*}
  y f'(xy) &= (y-1) e^{xy-x-y} \left( e^yf(x)+e^xf(y) \right) + e^{xy-x-y} \left( e^yf'(x)+e^xf(y) \right) \\
  &= e^{xy-x-y} \left( (y-1) e^yf(x)+(y-1) e^xf(y) + e^yf'(x)+e^xf(y) \right) \\
  &= e^{\color{red}{x}(y-1)-y} \left( e^yf'(x) + (y-1) e^yf(x)+y e^xf(y) \right)
\end{align*}
Now, I know why we got different calculations but the same result: replace $x$ and $y$ by $1$ and $x$ respectively, so the missing $\color{red}{x}$ doesn't affect the result, and we still have
\begin{equation*}
  f'(x)-f(x) = \frac{e^{x}}{x}.
\end{equation*}
A: On partial differentiating w.r.t $x$ $$f(xy)=e^{xy-x-y} \big( e^yf(x)+e^xf(y) \big)$$ 
we get 
$$yf'(xy)=e^{x(y-1)-y} (e^y f'(x)+e^y (y-1) f(x)+e^x y f(y))$$
Putting $x=1$ and $y=x$ we get :
$$\begin{align*}
  f'(x)-f(x) &= \frac{e^{x}}{x} \\
  f(x) e^{-x} &= \ln x + C \\
  f(x) &= e^{x} \ln x + C e^x \\
\end{align*}$$
Since $$ y = f(1) = 0, C = 0,$$
Hence $$y = e^x\ln(x).$$
