Consider $$\begin{aligned}\frac{d}{dt}x_{1} & =x_{2} \\ \frac{d}{dt}x_{2} & =-x_{1}-(\alpha+x_{1}^{2})x_{2}\end{aligned}$$
with $\alpha\ne 0$ and equilibrium $x^\ast=0$.
Then $f(x_{1},x_{2})=\begin{bmatrix}f_{1}(x_{1},x_{2}) \\ f_{2}(x_{1},x_{2})\end{bmatrix}=\begin{bmatrix}x_{2} \\ -x_{1}-(\alpha+x_{1}^{2})x_{2}\end{bmatrix}$
Thus we calculate the Jacobian as $$f'=\begin{bmatrix}\frac{\partial f_{1}}{\partial x_{1}} & \frac{\partial f_{1}}{\partial x_{2}} \\ \frac{\partial f_{2}}{\partial x_{1}} & \frac{\partial f_{2}}{\partial x_{2}}\end{bmatrix}=\begin{bmatrix} 0 & 1 \\ -1-2x_{1}x_{2} & -\alpha-x_{1}^{2}\end{bmatrix}$$
But I want to calculate $f'(x^{\ast})$ so that I can compute its eigenvalues and thus classify its stability. The problem I have is that I have one equilibrium point $x^{\ast}=0$, but two variables $x_{1}$ and $x_{2}$, so I'm a little confused with regards to the calculation.
Equilibrium/fixed/critical points, whatever you want to call them, are just the points $(x_1, x_2)$ where $\dot{x}_1 = 0$, $\dot{x}_2 = 0$. So find the values of $x_1$ and $x_2$ to satisfy this and these are your equilibrium coordinates.