Given $f(x)$ is a differentiable function such that $$f(x+y)=e^xf(y)+e^yf(x)$$ $\forall$ $x,y$ $\in$ $\mathbb{R}$ and $f'(0)=1$. Find $f(x)$
if we put $y=0$ we get
$$f(x)=e^xf(0)+f(x)$$ $\implies$ $$f(0)=0$$
if we put $x=-y$ we get $$f(0)=e^xf(-x)+e^{-x}f(x)$$ which $\implies$
$$e^xf(-x)=-e^{-x}f(x)$$ if we let $g(x)=e^{-x}f(x)$ we notice that $g$ is Odd function, so $g(0)=0$
Also given $f'(0)=1$ that is $$\lim_{h \to 0}\frac{f(h)}{h}=1$$ which can be written as $$\lim_{h \to 0}\frac{e^{-h}f(h)}{h}=\lim_{h \to 0}\frac{g(h)}{h}=1$$ Hence $g'(0)=1$
Now any hint from here to find $f(x)$
Given $$\displaystyle f(x+y) = e^{x}f(y)+e^{y}f(x)\Rightarrow \frac{f(x+y)}{e^{x+y}} = \frac{f(y)}{e^{y}}+\frac{f(x)}{e^x}$$
Now Put $\displaystyle \frac{f(x)}{e^x} = g(x)\;,$ Then functional equation convert into
$\displaystyle g(x+y) = g(x)+g(y)$
So it is a Cauchy functional equation whose solution is $g(x) = cx$
So we get $$\displaystyle \frac{f(x)}{e^x} = cx\Rightarrow f(x) = cxe^x$$
So we get $f'(x) = c\left[xe^x+e^x\right]$
Now given $f'(0) = 1$. So put $x=0$ in above equation, We get $1=c$
So we get $$f(x) = xe^x$$