What is the Continuity Equation in QM?

The other answers are correct but it's worth stating, given your title's question, what the continuity equation is actually telling us.

A continuity equation is the expression of balance between the rate of change of the amount of "stuff" inside a region $M$ on the one hand and the total flux of that stuff through the boundary $\partial M$ on the other. It is the translation into mathematics of the statement, "what goes in, stays in unless it comes out again through the boundary". For example, like a fluid: the amount of fluid in a volume can only change by the total flux of fluid through that volume's boundary. By shrinking the test volume and taking the limit, one can show that this notion is the same as the equation in Nate Stemen's answer if one takes $\rho$ to be the density of a fluid and $\vec{j}$ to be the mass flow rate.

In your case, the "stuff" is the total probability per unit volumne of the position operator's yielding a measurement within that volume. Probability flux is perhaps a little more abstract than mass flow rate, but the fulfilling of a continuity equation over the whole of space simply means that the probability that the measurement will be somewhere is constant. Which it is: the probability that the measurement will lie somewhere in all space is unity!! And the continuity equation results from applying this principle to an arbitrary volume, the volume's boundary and the volume's complement. The decrease in probability of measurement within the volume must match the increase in the probability of measurement without, which, in turn, must match the integrated flux through the boundary.

Now, ponder these thoughts for a bit, and with them in mind, see whether you can reproduce, from your own reasoning, AlphaGo's answer.

So the continuity equation is usually written as

$$\frac{\partial\rho}{\partial t} + \nabla\cdot \mathbf{j} = 0$$

where $\rho \equiv \rho(\mathbf{r},t) = |\psi|^2$ is the standard probability density and $\mathbf{j} = \frac{\hbar}{2mi}\left(\bar{\psi}\nabla\psi - \psi\nabla\bar{\psi}\right)$ is called the probability current. Just wanted to write this out so you can better understand what you are reading if you look online.

Now in your problem they have defined $\xi \equiv |\psi|^2 + \frac{\partial}{\partial x}(\bar{\psi}\psi)$ which is very close to $\rho + \mathbf{j}$ but not quite the same. What they want you to do is show $\partial_t\xi + \partial_x\xi = 0$