Explain why the solution curves of this system are orthogonal the level curves of $V(x,y)$ defined by $\nabla V(x,y)=(x',y')$ The system in question here is
$$
\begin{split}
\frac{dx}{dt} & = -2x(x-1)(2x-1) & = f(x,y)\\
\frac{dy}{dt} & = -2y &= g(x,y)
\end{split}
$$
The potential function,
$$V(x,y) = -x^4+2x^3-x^2-y^2+C_1$$
Because,
$$\nabla V(x,y)=(x',y')$$
To find the solution curves I wrote the solutions of the system in implicit form, such that the variables $x$ and $y$ are not dependent on $t$. This gave me
$$\frac{(x-\frac{1}{2})^2}{x(x-1)}\cdot y=C_2$$
We define a function $F$ as,
$$F(x,y)=\frac{(x-\frac{1}{2})^2}{x(x-1)}\cdot y=C_2$$
Asked is to show why the solution curves $(F(x,y)=\text{constant})$ are orthogonal to the level curves $V(x,y)=\text{constant}$. The formula for the solution curves was obtained by setting, $\frac{dy}{dx}=\frac{g(x,y)}{f(x,y)}$. The 2 curves are orthogonal if I can show that for the derivative of $y$ over $x$ of both the curves it holds that,
$$\frac{dy_1}{dx_1}\cdot\frac{dy_2}{dx_2}=-1$$
Thus I need to show that,
$$\frac{dy_2}{dx_2}=\frac{-f(x_2,y_2)}{g(x_2,y_2)}=\frac{\frac{\partial V(x_2,y_2)}{\partial x_2}}{\frac{\partial V(x_2,y_2)}{\partial y_2}}$$
Where $\frac{dy_2}{dx_2}$ corresponds to the derivative of the variables of $V$. I don't know how to show this, and I think deriving the function $F$ might have not been necessary. Since I want to use the derivative of the variable $y$ with respect to $x$, which I already have.
 A: The idea is the following:
Let us parametrize the level curve locally around a (smooth) point $P$ as $\gamma(t)$. Therefore, $V\circ \gamma=C=constant$.
We have to compute the tangent vector at that point, so let us derive both the members of our equation and compute it at $t=0$:
$$(V\circ \gamma)'(0)=0=\nabla V(\gamma(0))\cdot \gamma'(0).$$
But now, if you take the solution to your system and supposing that it is passing through $P$, then this implies that $\nabla V(\gamma(0))=V(P)=(x'(0),y'(0))$. Therefore we would get
$$(x'(0),y'(0))\cdot \gamma'(0)=0.$$
In other words the tangent vector at $P$ of the solution of your system is orthogonal to the tangent vector at $P$ of the level curve $V(x,y)=C$, that is what you are wanted to look for.
You can observe that this is an intrinsic property of the Potential and does not depend from your specific example.
Therefore, we can say in general that given a potential of a system, then the solution curve of the system is always orthogonal to the level curves of the Potential.
