Could someone explain getting a derivative when you have $y=f(x^2)$ added on So, if given a problem $y^3 + x^3 = 27$,
and the statement $y = f(x^2)$,
Could someone explain what it means to find $y'$? (using implicit differentiation) What does the $y = f(x^2)$ mean? Would this be equivalent to finding $dy/dx^2$? 
I'm just completely confused about this (and the usage of the chain rule... where how $v(x) = 1$) and would love some help. Sorry if this is too broad a question.
This isn't a textbook problem - our teacher wrote the two statements on the board and told us to "solve." - so I don't know the "exact" problem - is that problematic? Am I expected to be able to solve it like this?

If we solve the equation for $y$, we get $y=\sqrt[3]{27-x^3}$, but you are supposed to write it as a function of $x^2$, which we can call $u$.  $y=\sqrt[3]{27-u^{(\frac{3}{2})}}$, which works fine for $x\ge0$.  Presumably then you are asked to find $\frac{dy}{du}$.  This is a reasonable chain rule problem.
To do this implicitly, start with $x^3+y^3=27$, write it as  $u^{(\frac{3}{2})}+y^3=27$and take the derivative with respect to $u$ of each side.  You get  $\frac{3\sqrt{u}}{2}+3y^2\frac{dy}{du}=0$
