Answer
$f= \dfrac{ f_1 f_2 }{ \sqrt{ f_1^2 + f_2^2 }} $
Work Step by Step
We know that the force exerted on the spring is given by
$$F_{sp}=-kx$$
where $k$ is the spring constant and $x$ is the compressed or the stretched distance from the equilibrium point.
Now we can assume that these two springs are massless which means that the net force exerted on both of them is zero when we pull the block $m$ a distance of $x$ to the right.
When we pull the block a distance $x$ to the right, the first spring will be stretched by a distance of $x_1$ while the second spring will be stretched by a distance $x_2$. See the figure below.
So,
$$x_{block}=x_1+x_2\tag 1$$
when pulling the block to the right then the attached spring to it (which is spring 2) will be affected by a force of $F_{\rm block\; on\;2}=-k_2x_2$ that pulls it to the right. And since the net force exerted on the spring is zero, so the force exerted on spring 2 by spring 1 must have the same magnitude.
So,
$$F_{\rm 1\;on\;2}=F_{\rm block\; on\;2}=k_2x_2\tag 2$$
Now the wall pulls spring 1 back by a force that must equal to the spring force exerted on it by spring 1 (according to Newton's third law).
Also, the net force exerted on spring 1 is zero, so
$$F_{\rm 2\;on\;1}=F_{\rm wall\; on\;1}=k_1x_1\tag3$$
From (2) and (3), we got that
$$F_{\rm block}=k_1x_1=k_2x_2\tag4$$
So, $x_1=F_{\rm block}/k_1$, and $x_2=F_{\rm block}/k_2$ plug these into (1).
$$x_{block}=\dfrac{F_{\rm block}}{k_1}+\dfrac{F_{\rm block}}{k_2}$$
$$x_{block}=F_{\rm block}\left[\dfrac{1}{k_1}+\dfrac{1}{k_2}\right]$$
$$x_{block}=F_{\rm block}\left[ \dfrac{k_1+k_2}{k_1k_2}\right]\tag 5$$
Now if we replaced these two springs with one effective spring and pulled the same block attached to it to the right by the same distance $x_{block}$, the force magnitude exerted by the block on the spring is now given by
$$F_{\rm block}=k_{\rm effective}x_{\rm block}$$
Thus, $x_{\rm block}=F_{\rm block}/k_{\rm effective}$
Plugging into (5);
$$\dfrac{\color{red}{\bf\not} F_{\rm block}}{k_{\rm effective}}=\color{red}{\bf\not} F_{\rm block}\left[ \dfrac{k_1+k_2}{k_1k_2}\right] $$
Thus,
$$k_{\rm effective}= \dfrac{k_1k_2}{k_1+k_2}\tag 6$$
Now if we let the system oscillate freely, the angular frequency of the block is then given by
$$\omega=\sqrt{\dfrac{k_{\rm effective}}{m}}$$
Plugging from (6);
$$\omega=2\pi f=\sqrt{\dfrac{ k_1k_2}{m(k_1+k_2)}}$$
Hence,
$$f=\dfrac{1}{2\pi }\sqrt{\dfrac{ k_1k_2}{m(k_1+k_2)}}\tag 7$$
Now if each spring is acting alone on the block, then the angular frequency of each one is given by
$$\omega_1=\sqrt{\dfrac{k_1}{m}}\;\;\;,\;\;\;\;\omega_2=\sqrt{\dfrac{k_2}{m}}$$
Hence,
$$\omega_1^2= \dfrac{k_1}{m} \;\;\;,\;\;\;\;\omega_2^2= \dfrac{k_2}{m} $$
$$k_1=m\omega_1^2 \;\;\;,\;\;\;\; k_2=m \omega_2^2 $$
Plugging these into (7);
$$f=\dfrac{1}{2\pi }\sqrt{\dfrac{ (m\omega_1^2 )(m\omega_2^2 )}{m[(m\omega_1^2 )+(m\omega_2^2 )]}} $$
$$f=\dfrac{1}{2\pi }\sqrt{\dfrac{ \color{red}{\bf\not} m^2\omega_1^2 \omega_2^2 }{ \color{red}{\bf\not} m^2\omega_1^2 + \color{red}{\bf\not} m^2\omega_2^2 }} $$
$$f=\dfrac{1}{2\pi }\sqrt{\dfrac{ \omega_1^2 \omega_2^2 }{ \omega_1^2 + \omega_2^2 }} $$
And since, $\omega=2\pi f$, then $\omega_1=2\pi f_1$, and $\omega_2=2\pi f_2$
$$f=\dfrac{1}{2\pi }\sqrt{\dfrac{ (2\pi f_1)^2 (2\pi f_2)^2 }{ (2\pi f_1)^2 + (2\pi f_2)^2 }} $$
$$f=\dfrac{1}{2\pi }\sqrt{\dfrac{ 16\pi^4 f_1^2 f_2^2 }{ 4\pi^2 [f_1^2 + f_2^2] }} $$
$$f=\dfrac{1}{2\pi }\sqrt{\dfrac{ 4\pi^2 f_1^2 f_2^2 }{ f_1^2 + f_2^2 }} $$
$$f=\dfrac{\color{red}{\bf\not} 2\color{red}{\bf\not} \pi}{\color{red}{\bf\not} 2\color{red}{\bf\not} \pi }\sqrt{\dfrac{ f_1^2 f_2^2 }{ f_1^2 + f_2^2 }} $$
$$\boxed{f= \dfrac{ f_1 f_2 }{ \sqrt{ f_1^2 + f_2^2 }}} $$
