How would I differentiate y = 3xy + 2x^2 + x^2y^2 ?

Since this problem contains both xs and ys on the right hand side, we need to use implicit differentiation. This is where we use the chain rule to differentiate with regards to x the terms which contain ys. For example, if we wanted to differentiate y² with regards to x, we would first differentiate with regards to y, giving 2y, and then consider how to adjust that answer. Essentially what we have done here is d/dy(y²), but what we want to do is d/dx(y²). In order to change our answer for d/dy to an answer for d/dx we must multiply by dy/dx. You can see that the dys cancel and leave d/dx, which is what we want. Therefore the answer to d/dx(y²) will be 2y * dy/dx.Now, let's use this on the original question. We will handle the equation one term at a time. Starting with the left hand side, the y differentiates to dy/dx. Moving onto the right hand side, to differentiate 3xy we need to use the product rule, so we do the first product (3x) times the differential of the second (dy/dx) plus the second (y) times the differential of the first (3), giving 3x * dy/dx + 3y. The next term (2x²) is a simple multiply by the power and reduce the power by one to give 4x. Finally the last term (x²y²) is also a product, so we will again use the product rule. We must do the first product times the differential of the second plus the second times the differential of the first. The first product is x². The differential of the second product must be found using implicit differentiation - so start by working out d/dy(y²) and then multiply by dy/dx to adjust like we did in the earlier example. This gives us 2y * dy/dx, which we are multiplying by x²(the first product) to give 2x²y * dy/dx. Then the second product (y²) is multiplied by the differential of the first (2x - another simple multiply by the power and reduce the power by one) to give 2xy². Putting this all together we get dy/dx = 3x * dy/dx + 3y + 4x + 2x²y * dy/dx + 2xy².