$\because$ $f(x + y) = {2^x}f(y) + {4^y}f(x)$ ....... (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 $k = {1 \over 4}$ $\therefore$ $f(x) = {{{4^x} - {2^x}} \over 4}$ $\therefore$ $f'(x) = {{{4^x}\ln 4 - {2^x}\ln 2} \over 4}$ $f'(x) = {{({{2.4}^x} - {2^x})ln2} \over 4}$ $\therefore$ ${{f'(4)} \over {f'(2)}} = {{2.256 - 16} \over {2.16 - 4}}$ $\therefore$ $14{{f'(4)} \over {f'(2)}} = 248$