Function $x(t)=e^{j\pi t}\cos(\frac{2\pi t}{3})=[\cos(\pi t)-\sin(\pi t)]\cos(\frac{2\pi t}{3})$.
j is i, the imaginary number, square root of -1.
This is the graph:
EDIT: I did it. But not sure if I did it right.
So $x(t)=e^{-j\pi t}cos(\frac{2\pi}{3}t)$. It is period only if we can find a T such that x(t)=x(t+T). So $$e^{-j\pi t}cos(\frac{2\pi}{3}t)=e^{-j\pi t}e^{-j\pi T}cos(\frac{2\pi}{3}t+\frac{2\pi}{3}T)$$. We need to find a T so that $e^{-j\pi T}=1$ and $cos(\frac{2\pi}{3}t+\frac{2\pi}{3}T)=cos(\frac{2\pi}{3}t)$.
For the second condition (that the two cosines must equal), T=3k, where k is a natural number. For the first condition, using Euler's theorem, $\pi T=2\pi k$ because then cosine part of the euler's equation is 1 and the sine part is 0.
So you have T=2k, for first condition, and T=3k, for the second. So, the combine T=6k, where k=1,2,3...
Hint: the product of two periodic functions with periods $\,2\,$ and $\,3\,$, respectively, has period $6$.