You're a student, so I'll try to help you figure out the solution by yourself. Make sure you understand the following points:
- The signal is a triangle of length $2N+1$.
- A triangular sequence can be written as the convolution of two even more basic sequences, each of length $N+1$.
- Convolution in the time domain corresponds to multiplication of the $\mathcal{Z}$-transforms.
- Define the basic signal determined in point 2. in the interval $0\le n\le N$ and compute its $\mathcal{Z}$-transform.
- Take the square of that $\mathcal{Z}$-transform, which gives you the $\mathcal{Z}$-transform of a triangle in the range $0\le n\le 2N$.
- In order to get the transform of $x[n]$, you need to shift the triangle to the left by $N$, which corresponds to a multiplication by $z^N$.
If you did everything right, the resulting $\mathcal{Z}$-transform should be
$$X(z)=\left(\frac{1-z^{N+1}}{1-z^{-1}}\right)^2z^N\tag{1}$$
You obtain the Fourier transform by substituting $z=e^{j\omega}$ in $(1)$. Note that the result must be real-valued due to the symmetry of $x[n]$. I leave the intermediate steps up to you, but your solution should be
$$X(e^{j\omega})=\left(\frac{\sin\left(\frac{(N+1)\omega}{2}\right)}{\sin\left(\frac{\omega}{2}\right)}\right)^2\tag{2}$$