Now on the left-hand side, we of course are going to get the second derivative of y with respect to x. But what do we get on the right-hand side? And there's multiple ways to approach this, but for something like this, the quotient rule probably is the best way to tackle it. I sometimes complain about the quotient rule, saying, hey, it's just a variation of the product rule, but it's actually quite useful in something like this. We just have to remind ourselves that this is going to be equal to the derivative of the numerator with respect to x, and so that's just going to be derivative of y with respect to x times the denominator, three y squared minus x, minus the numerator, y, times the derivative of the denominator with respect to x. Well, what's the derivative of this denominator with respect to x? Well, the derivative of three y squared with respect to x, that's going to be the derivative of three y squared with respect to y, which is just going to be six y, I'm just using the power rule there, times the derivative of y with respect to x.