Derivative of a Lyapunov function for a nonlinear system 
Let
$$\begin{aligned}\begin{cases}\dot{x}_{1}=-\left( 2x_{1}-x_{2}\right)^3+\left( x_{1}-x_{2}\right)  \\
\dot{x}_{2}= -\left( 2x_{1}-x_{2}\right) ^{3}+2\left( x_{1}-x_{2}\right)\end{cases}\\
 \end{aligned}$$
Given
$$V\left( \mathbb{x}\right) =\mathbb{x}^{T}P\mathbb{x} \qquad P=\begin{bmatrix}
5 & -3 \\
-3 & 2
\end{bmatrix}$$
Answer: $\dot V\left( x_{1},x_{2}\right) =-2\left[ \left( 2x_{1}-x_{2}\right) ^{4}+\left( x_{1}-x_{2}\right) ^{2}\right]$

My attempt:
We can write $V(\mathbb{x})$ as
$$\begin{aligned}V\left( x_1,x_2\right) &=5x_{1}^{2}-6x_{1}x_{2}+2x_{2}^{2}\\
&=\left( 2x_{1}-x_{2}\right) ^{2}+\left( x_{1}-x_{2}\right) ^{2}\end{aligned}$$
and defining the change of variables $z_{1}=2x_{1}-x_{2}$ and $z_{2}=x_{1}-x_{2}$ we obtain
$$V(z_1,z_2)=z_1^2+z_2^2$$
and
$$\begin{cases}\dot{z}_{1}=-z_{1}^{3}+z_{2}\\
\dot{z}_{2}=-z_{1}^{3}+2z_{2}\end{cases}$$
Now we can calculate $\dot V(z_1,z_2)$ as
$$\begin{aligned}\dot V\left( z_{1},z_{2}\right) &=\dfrac{\partial V}{\partial z_{1}}\left( z_{1},z_{2}\right) \dot{z}_{1}+\dfrac{\partial V}{\partial z_{2}}\dot{z}_{2}\\
&=2z_{1}\left( -z_{1}^{3}+z_{2}\right) +2z_{2}\left( -z_{1}^{3}+2z_{2}\right) \end{aligned}$$
 A: This is incorrect:
$$\begin{cases}\dot{z}_{1}=-z_{1}^{3}+z_{2}\\
\dot{z}_{2}=-z_{1}^{3}+2z_{2}\end{cases}$$
It should be like this:
$$
\dot z_1= 2\dot x_{1}-\dot x_{2}= 
2(-\left( 2x_{1}-x_{2}\right)^3+\left( x_{1}-x_{2}\right))
-(-\left( 2x_{1}-x_{2}\right) ^{3}+2\left( x_{1}-x_{2}\right))
$$
$$
=2(-z_1^3+z_2)+z_1^3-2z_2= -z_1^3
$$
$$
\dot z_2= \dot x_1-\dot x_2= -\left( 2x_{1}-x_{2}\right)^3+\left( x_{1}-x_{2}\right)
-(-\left( 2x_{1}-x_{2}\right) ^{3}+2\left( x_{1}-x_{2}\right))
$$
$$
=-z_1^3+z_2+z_1^3-2z_2= -z_2
$$
A: In addition to AVK's answer, I just wanted to point out how for this type of problem you can usually find $\dot{V}(x)$ by leveraging the symmetry of $P$. This sometimes saves a lot of busy work:
First, notice that $V(x)$ can be written as a scalar product:
$$V(x)=\langle x,Px\rangle$$
Hence:
$$\begin{aligned}\dot{V}(x)&=\langle \dot{x},Px\rangle +\langle x,P\dot{x}\rangle \\&= \langle P^T\dot{x},x\rangle +\langle x,P\dot{x}\rangle \\&= \langle x,P^T\dot{x}\rangle +\langle x,P\dot{x}\rangle\end{aligned}$$
And thanks to $P=P^T$ we obtain:
$$\dot{V}(x)=2\langle x,P\dot{x}\rangle = 2x^TP\dot{x}$$
To demonstrate this technique on your case, using $A=-(2x_1-x_2)^3$ and $B=(x_1-x_2)$ as shorthands, we get:
$$\begin{aligned}\dot{V}(x)&=2x^T\left(\begin{array}{cc}5 & -3\\ -3 & 2\end{array}\right)\left(\begin{array}{c}A+B \\ A+2B\end{array}\right) \\ &=2x^T\left(\begin{array}{c}2A-B\\ -A+B\end{array}\right)\\&=2\left(x_1(2A-B)-x_2(A-B)\right) \\ &=2\left((2x_1-x_2)A-(x_1-x_2)B\right) \\&=
-2\left( \left( 2x_{1}-x_{2}\right) ^{4}+\left( x_{1}-x_{2}\right) ^{2}\right)\end{aligned}$$
I just find the calculation of $\dot{V}(x)$ to be more straightforward this way in general !
