Answer
$f = \sqrt{f_1^2+f_2^2}$
Work Step by Step
If the mass is a distance of $x$ from the equilibrium point, the force on the mass is $k_1x+k_2x$ which is $(k_1+k_2)x$. Thus, the equivalent spring constant is $k_1+k_2$. We can find an expression for the frequency using:
$f = \frac{1}{2\pi}~\sqrt{\frac{k_1+k_2}{m}}$
We can find an expression for $\sqrt{f_1^2+f_2^2}$ using the formula above:
$\sqrt{f_1^2+f_2^2}$
$= \sqrt{(\frac{1}{2\pi}~\sqrt{\frac{k_1}{m}})^2+ (\frac{1}{2\pi}~\sqrt{\frac{k_2}{m}})^2}$
$= \sqrt{(\frac{1}{2\pi})^2~\frac{k_1}{m}+ (\frac{1}{2\pi})^2~\frac{k_1}{m}}$
$= \sqrt{(\frac{1}{2\pi})^2~\frac{k_1+k_2}{m}}$
$= \frac{1}{2\pi}~\sqrt{\frac{k_1+k_2}{m}}$
Therefore, $f = \sqrt{f_1^2+f_2^2}$.