The question is to find the zeros of $h(z)=\frac{1}{3}e^z-z$ inside $\mid z \mid \leq 1$
Preparing for an application of Rouche's theorem with $f(z)=\frac{1}{3}e^z$ and $g(z)=-z$ it is easy to see that $\mid g(z) \mid=\mid e^{i \theta} \mid=1$
However, I'm not sure how to investigate the behavior of $\mid f(z) \mid $ on the boundary of the unit disc, which is a usual problem I run into in applications of Rouche's Theorem.
$\mid f(z) \mid = \frac{1}{3}\mid e^z \mid=?$
Plugging in $z=e^{i\theta}$ is the only tool I have, and it doesn't seem to be useful here.
If $z$ fulfills $|z|\leq 1$ then $\left|\frac{1}{3}e^z\right| = \frac{1}{3} e^{\text{Re}(z)} \leq \frac{e}{3} < 1$, in particular on the boundary of the unit disk the function $\frac{1}{3}e^{z}$ is always smaller that $z$ (or $-z$) in absolute value. It follows that the number of zeroes of $h(z)$ inside the unit disk is the same as the number of zeroes of $-z$ inside the unit disk, i.e. $\color{red}{\large 1}$. Such zero is real and belong to the interval $\left(\frac{1}{2},\frac{3}{4}\right)$.