I am trying to find the linearisation of $$x'''(t)+2x'(t)+x^2(t)=0 \tag{1},$$ about the initial point $\underline{x}(0)=0$.
My attempt:
I first convert $(1)$ into a system of first-order ODEs. We let $x_1=x, \ x_2=x'$ and $x_3=x''$. Then, $$\underline{x}=\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \ \underline{x'}=\begin{pmatrix} x_2 \\ x_3 \\ -(2x_2+x_1^2) \end{pmatrix} \ \text{with initial conditions} \ \ \underline{x}(0)=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.$$ To linearise the system, we consider $\ \underline{x'}=\underline{f}(\underline{x}(0))+\underline{f_x}(\underline{x}(0))(\underline{x}-\underline{x}(0))$, where $\underline{f_x}$ denotes the Jacobian. Now, $$\underline{f}(\underline{x}(0))=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}, \ \underline{f_x}(\underline{x}(0))=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1\\ 0 & -2 & 0 \end{pmatrix}.$$ Putting this together, $$\underline{x'}=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1\\ 0 & -2 & 0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}.$$
Your solution is correct. BTW, in this case you can simply throw away the $x^{2}$ term to get the linearized equation (since your $f$ is already polynomial around $\boldsymbol{x}=0$ and the expansion is unique).