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 ______

Given

$$f(x+y)=2^x f(y)+4^y f(x).$$

Putting

$$x=y=0,$$

we get

$$f(0)=f(0)+f(0),$$

which gives

$$f(0)=0.$$

Now put

$$y=0.$$

Then

$$f(x)=2^x f(0)+f(x),$$

which is consistent since

$$f(0)=0.$$

Next, put

$$y=2.$$

Then

$$f(x+2)=3\cdot 2^x+16f(x).$$

Differentiating,

$$f'(x+2)=3(\ln 2)2^x+16f'(x).$$

Putting

$$x=2,$$

$$f'(4)=12\ln2+16f'(2). \qquad (1)$$

Now put

$$x=y.$$

Then

$$f(2x)=\left(2^x+4^x\right)f(x).$$

Differentiating,

$$2f'(2x)=\left[(\ln2)2^x+(\ln4)4^x\right]f(x)+(2^x+4^x)f'(x).$$

Putting

$$x=2,$$

$$2f'(4)=\left(4\ln2+32\ln2\right)\cdot 3+20f'(2).$$

Hence

$$2f'(4)=108\ln2+20f'(2). \qquad (2)$$

Substituting (1) into (2),

$$2(12\ln2+16f'(2))=108\ln2+20f'(2).$$

Thus

$$24\ln2+32f'(2)=108\ln2+20f'(2),$$

which gives

$$12f'(2)=84\ln2,$$

$$f'(2)=7\ln2.$$

From (1),

$$f'(4)=12\ln2+16(7\ln2)=124\ln2.$$

Therefore,

$$14\cdot\frac{f'(4)}{f'(2)}=14\cdot\frac{124\ln2}{7\ln2}=2\cdot124=248.$$