It is given that $f$ is a monotonically increasing function $\forall x\in\mathbb R$ such that $f''(x) \gt 0$.
Provided that $f^{-1}$ exists, It is to be shown that, $$\frac{f^{-1}(x_1) + f^{-1}(x_2) +f^{-1}(x_3)}{3} \lt f^{-1}\left(\frac{x_1 +x_2 +x_3}{3}\right)$$
I assumed that $f^{-1}(x)$ is the mirror image of $f(x)$ about the origin, it must have all values of its derivatives negative, since $f$ is monotonically increasing. Is this correct? If not, how can the above be proved?
The inequality you want is a consequence of Jensen's inequality.
Since $f$ is strictly convex we get that
$$\frac{f(y_1) + f(y_2) +f(y_3)}{3} >f(\frac{y_1 + y_2+y_3}{3})$$
Now set $y_i=f^{-1}(x_i)$
So the previous transforms to
$$\frac{x_1 + x_2 +x_3}{3} >f\left (\frac{f^{-1}(x_1)+f^{-1}(x_2)+f^{-1}(x_3)}{3}\right )$$
Since $f$ is increasing so is $f^{-1}$ so applying $f^{-1}$ to the previous inequality we obtain the desired inequality namely
$$f^{-1} \left (\frac{x_1 + x_2 +x_3}{3} \right ) > \frac{f^{-1}(x_1)+f^{-1}(x_2)+f^{-1}(x_3)}{3}$$