I'm going to have to do implicit differentiation one way or the other. So what if I multiplied both sides of this times three y squared minus x? Then I get three y squared minus x times dy dx is equal to y. And so now taking the derivative of both sides of this is going to be a little bit more straightforward. In fact, if I want, I could, well actually, let me just do it like this. And so let's apply our derivative operator to both sides. Let's give ourselves some space.