Suppose we have two curves that cross at right angles. We represent them in polar coordinates as functions $r_1$ and $r_2$ from angle to radius. Suppose the crossing happens at $(r, \theta)$, so that $r_1(\theta) = r_2(\theta) = r$.
For $i \in {1, 2}$, put $x_i(\theta) = r_i(\theta)\cos(\theta)$ and $y_i(\theta) = r_i(\theta)\sin(\theta)$. Then $\frac{dx_i}{d\theta} = \frac{dr_i}{d\theta}\cos(\theta) - r_i\sin(\theta)$, and $\frac{dy_i}{d\theta} = \frac{dr_i}{d\theta}\sin(\theta) + r_i\cos(\theta)$.
The curves cross at right angles, so we have $\frac{dx_1}{d\theta}\frac{dx_2}{d\theta} + \frac{dy_1}{d\theta}\frac{dy_2}{d\theta} = 0$. (This can be deduced from $\frac{dy_1}{dx_1} = -\frac{dx_2}{dy_2}$ when they are finite, but it is more general). Substitute:
$$
\left(\frac{dr_1}{d\theta}\cos(\theta)-r\sin(\theta)\right)
\left(\frac{dr_2}{d\theta}\cos(\theta)-r\sin(\theta)\right) +
\left(\frac{dr_1}{d\theta}\sin(\theta)+r\cos(\theta)\right)
\left(\frac{dr_2}{d\theta}\sin(\theta)+r\cos(\theta)\right) = 0
$$
Multiply out the brackets, cancel the terms in $\cos(\theta)\sin(\theta)$, and use $\cos^2(\theta) + \sin^2(\theta) = 1$:
$$\frac{dr_1}{d\theta}\frac{dr_2}{d\theta} + r^2 = 0$$
Divide through by $r^2$ and subtract 1:
$$\left(\frac1r\frac{dr_1}{d\theta}\right) \left(\frac1r\frac{dr_2}{d\theta}\right) = -1$$
If $x$ is playing the role of the position and $y$ is the momentum, Hamilton's equations should read $$x' = \frac{\partial H}{\partial y}\quad \mbox{and}\quad y' = -\frac{\partial H}{\partial x}.$$Can you find a function $H = H(x,y)$ such that $$\frac{\partial H}{\partial x} = x-x^3 \quad\mbox{and}\quad \frac{\partial H}{\partial y} = 2y^3-y?$$Clearly $$H(x,y) = \frac{x^2}{2} - \frac{x^4}{4} + \frac{y^4}{2} - \frac{y^2}{2} = \frac{1}{2}(x^2-y^2 + y^4) - \frac{x^4}{4}$$does the job. Conservation of energy says that once you solve the system for $x(t)$ and $y(t)$, it will turn out that $$\frac{{\rm d}}{{\rm d}t} H(x(t), y(t)) = 0$$and thus $H(x(t),y(t)) = H(x_0,y_0)$ for all $t$, where $(x_0,y_0)$ are the initial conditions given for the system. Say that you fix a real number $a \in \Bbb R$. Each inverse image $H^{-1}(a)$ will be an orbit for your system. So the point is, can you graph all the curves $$\frac{1}{2}(x^2-y^2 + y^4) - \frac{x^4}{4} = a,$$as $a$ ranges over $\Bbb R$? This will be your phase portrait. The upshot is that once you have recognized $H$, you can (in principle) draw this without solving the actual system.
Best Answer
The phase plane equation for the integral curves is
$$\def\part#1#2{\frac{\mathrm d#1}{\mathrm d#2}} \part yx=\frac{\part yt}{\part xt}=\frac{\mathrm e^x+y}{2y-x}\;.$$
To get a grip on this, it helps to expand around the equilibrium. At equilibrium, $\part xt=2y-x=0$, and thus $y=x/2$, so we can write $y=x/2+z$, with $y'=1/2+z'$. Substituting this into the differential equation for $y$ yields
$$ \begin{align} \frac12+z' &=\frac{\mathrm e^x+\frac x2+z}{2(\frac x2 + z)-x}\\ &=\frac{\mathrm e^x+\frac x2+z}{2z}\;,\\ z' &=\frac{\mathrm e^x+\frac x2}{2z}\;,\\ 2zz' &=\mathrm e^x+\frac x2\;,\\ z^2 &=\mathrm e^x+\frac {x^2}4+C\;,\\ z &=\pm\sqrt{\mathrm e^x+\frac {x^2}4+C}\;,\\ y &=\frac x2\pm\sqrt{\mathrm e^x+\frac {x^2}4+C}\;. \end{align}$$