I have a vector field $\frac{dx}{dt}=y, \frac{dy}{dt}=(-x^3+x^2-2x)+y(b-x^2)$.
I have found that when $b=x^2$, the Jacobian of the system at $(x,0)$ has pure imaginary eigenvalues, so there should be Hopf bifurcations at such values of $b$. I found $x=\frac{1}{4}, \frac{25}{4}$ because $x$ should be a root of $(-x^3+x^2-2x)$. Then I started plotting and only found very uninteresting looking phase planes.
Can someone point out what my problem might be?
OK, it seems that you don't really have much of an understanding of what is going on here. Perhaps you might need to have a bit more a read up before attempting problems such as these? Anyhow, I will now go through the steps that you should have taken.
Initially the equilibria $(x_e, y_e)$ of the system \begin{align} \frac{d x}{d t} & = y & \equiv f_1(x,y,b), \\ \frac{d y}{d t} & = -x^3 + x^2 - 2x + y(b - x^2) & \equiv f_2(x,y,b), \end{align} need to be found. These are identified through solving \begin{align} f_1(x_e,y_e,b) & = 0, \\ f_2(x_e,y_e,b) & = 0. \end{align}
Upon doing this we find that the system only has one equilibrium at $(x_e, y_e) = (0,0)$. The next step is to linearize the system about this equilibrium point, we do this through the use of the Jacobian, J. The eigenvalues $\lambda$ of J determine the stability of this equilibrium point. The eigenvalues of J satisfy the quadratic \begin{equation} \lambda^2 - b \lambda + 2 = 0. \end{equation}
Now we solve for the roots of this quadratic to find the eigenvalues and we consider their behaviour for general values of $b$. In particular, we see that there is a qualitative change in the nature of the equilibrium point as $b$ passes through zero, that is, its stability changes. The gist of it is that for $b < 0$ the equilibrium point is stable, and for $b > 0$ it is unstable. And we also note that at $b = 0$ the real part of both eigenvalues is zero.
Now the linearization theorem does not hold when the real part of an eigenvalue is zero, so we suspect that this is a point of bifurcation. In this case because we only have one equilibrium point for all $b$ we suspect that the bifurcation is of Hopf type, and we also suspect this because both eigenvalues have zero real part (as opposed to just one of them). So we predict that a Hopf bifurcation gives rise to a stable limit cycle for $b > 0$.
The easiest way to determine if a limit cycle is actually born is to numerically solve the ODE system for a value of $b$ where we expect it to be. Below I've shown a phase plane portrait for a couple of values of $b$.
So we see that for $b < 0$ the equilibrium is indeed have a stable, and for $b > 0$ it is unstable and is surrounded by a stable limit cycle.