$$\because$$ $$f(x + y) = {2^x}f(y) + {4^y}f(x)$$ ....... (1)

Now, $$f(y + x){2^y}f(x) + {4^x}f(y)$$ ...... (2)

$$\therefore$$ $${2^x}f(y) + {4^y}f(x) = {2^y}f(x) + {4^x}f(y)$$

$$({4^y} - {2^y})f(x) = ({4^x} - {2^x})f(y)$$

$${{f(x)} \over {{4^x} - {2^x}}} = {{f(y)} \over {{4^y} - {2^y}}} = k$$

$$\therefore$$ $$f(x) = k({4^x} - {2^x})$$

$$\because$$ $$f(2) = 3$$ then $$k = {1 \over 4}$$

$$\therefore$$ $$f(x) = {{{4^x} - {2^x}} \over 4}$$

$$\therefore$$ $$f'(x) = {{{4^x}\ln 4 - {2^x}\ln 2} \over 4}$$

$$f'(x) = {{({{2.4}^x} - {2^x})ln2} \over 4}$$

$$\therefore$$ $${{f'(4)} \over {f'(2)}} = {{2.256 - 16} \over {2.16 - 4}}$$

$$\therefore$$ $$14{{f'(4)} \over {f'(2)}} = 248$$