Solve the functional Equation:- $f(x+y)=3^xf(y)+9^yf(x)\forall x,y\in \mathbb{R}$ 
Problem Statement:-
Consider a differentiable function $f:\mathbb{R}\rightarrow\mathbb{R}$ for which $f(1)=6$ and $$f(x+y)=3^x\cdot f(y)+9^y\cdot f(x),\;\;\forall x,y\in \mathbb{R}$$ then find $f(x)$.

My attempt:-
As the function is differentiable hence, we can differentiate the function w.r.t $x$ and $y$ but since $x$ and $y$ are independent, so $\dfrac{dy}{dx}$, so we get
$$f'(x+y)=3^x\ln3\cdot f(y)+9^y\cdot f'(x)\tag{1}$$
From the functional equation given in the problem we have
$$f(0+0)=3^0\ln{3}\cdot f(0)+9^0\cdot f(0)\implies f(0)=0$$
$$\therefore (1)\implies f'(y)=\ln{3}f(y)+c\cdot 9^y\tag{where $c=f'(0)$}$$
I am pretty much stuck after this.

If possible can you also tell me the line of thought that I should have while solving these types of problems.


I just happened to come up with an answer that doesn't need any differentiation.
We are given 
$$f(x+y)=3^x\cdot f(y)+9^y\cdot f(x)$$
On interchanging $x$ and $y$, we get 
$$f(y+x)=3^y\cdot f(x)+ 9^x\cdot f(y)$$
we get $$3^x\cdot f(y)+9^y\cdot f(x)=3^y\cdot f(x)+ 9^x\cdot f(y)\\
\implies f(y)\cdot (3^x-9^x)=f(x)\cdot (3^y-9^y)\\
\implies \dfrac{f(x)}{3^x-9^x}=\dfrac{f(y)}{3^y-9^y}=k(\text{say})$$
So, we get 
$$f(x)=k({3^x-9^x})$$
Since $f(1)=6$, so 
$$f(1)=k(-6)=6\implies k=-1$$
$$\therefore \boxed{f(x)=9^x-3^x}$$
I don't know whether there are some other functions that satisfy the given conditions, if there are then please do post that in your answer. 

And if you have some other approach please do post so that there is something new to learn.

