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.