Let $$f : \mathbb{R} \to \mathbb{R}$$ satisfy $$f(x + y) = 2^x f(y) + 4^y f(x), \forall x, y \in \mathbb{R}$$. If $$f(2) = 3$$, then $$14 \cdot \frac{f'(4)}{f'(2)}$$ is equal to ______

First, by setting $$x=y=0 in the given equation we obtain f(0)=f(0)+f(0)\implies f(0)=0$$.

Next, setting $$y=0 yields f(x)=2^x f(0)+f(x)=f(x)\,, $$ which is trivially true.

Now, setting $$x=0 gives f(y)=f(y)+4^y f(0)=f(y)\,, $$ also trivially true.

Since these identities hold, we try the form $$f(x)=k(4^x-2^x). Then f(x+y)=k(4^{x+y}-2^{x+y})=k(4^x\cdot4^y-2^x\cdot2^y)\,, while the right-hand side becomes 2^x f(y)+4^y f(x)=2^x\cdot k(4^y-2^y)+4^y\cdot k(4^x-2^x) =k(2^x\cdot4^y-2^x\cdot2^y+4^y\cdot4^x-4^y\cdot2^x) = k(4^{x+y}-2^{x+y})\,, $$ confirming the guess.

Substituting $$f(2)=3 into f(x)=k(4^x-2^x) gives k(16-4)=12k=3\implies k=\frac14\,, hence f(x)=\frac{4^x-2^x}{4}\,.$$

Therefore the derivative is $$f'(x)=\frac{4^x\ln4-2^x\ln2}{4}\,. It follows that f'(4)=$$ $$\frac{256\ln4-16\ln2}{4}=\frac{256\cdot2\ln2-16\ln2}{4}=\frac{496\ln2}{4}$$ $$=124\ln2 and f'(2)=$$ $$\frac{16\ln4-4\ln2}{4}=\frac{32\ln2-4\ln2}{4}=\frac{28\ln2}{4}$$ $$=7\ln2\,.$$

Finally, $$\frac{f'(4)}{f'(2)}=\frac{124\ln2}{7\ln2}=\frac{124}{7} and so 14\cdot\frac{f'(4)}{f'(2)}=14\cdot\frac{124}{7}=2\times124=248\,, giving the answer 248$$.