Basic differentiation: second derivative I'm currently teaching myself some differential equations by watching the MIT OCW series on the topic. In This video, at 21:50mins, the lecturer calculates the following derivatives:
1st $y'=x^2-y^2$
2nd $y''=2x-2yy'$
My simple question is, how he came to the second one. Is this a "total derivative" and why is it required? If I try to calculate the total derivative of y', I get:
$y''=(2x-1)dx+(1-2y)dy=2x dx - 2y dy$
I'm pretty sure that I made a silly mistake. Thanks for your help!

What's going on in the video, and in your posted problem, is what we refer to as implicit differentiation.
We view $y$ as a function of $x$, and thus, need to use the chain rule: $$y'(x) = x^2 - [y(x)]^2 \implies y''(x) = 2x - 2y(x)y'(x)$$  The author simply omits the parenthetical argument $(x)$: $$y'' = 2x - 2yy''$$
You can also see that your suggested answer just stops short one step: divide your $y''$ through by $dx$ and simplify! So you are not wrong; you "simply" haven't simplified!
