jmc

Switching the roles of $x$ and $y,$ we get $$f(y+x)=3^{x}f(y)+2^{y}f(x).$$Hence, $$3^{y}f(x)+2^{x}f(y)=3^{x}f(y)+2^{y}f(x).$$Then $$(3^y-2^y)f(x)=(3^x-2^x)f(y),$$so for $x\ne 0$ and $y\ne 0,$ $$\frac{f(x)}{3^x-2^x}=\frac{f(y)}{3^y-2^y}.$$Setting $y=1,$ we get $$\frac{f(x)}{3^x-2^x}=\frac{f(1)}{3^1-2^1}=1,$$so $f(x)=\boxed{3^x-2^x}.$ Note that this formula also holds for $x=0.$