You can either use Lagrange multipliers, or adopting your approach, on the boundary, note that $y^2 = 1-x^2$, so your function becomes $4x^3 + 3(1-x^2)$, can you find the extrema?
Compare the boundary extrema to the ones inside.
UPD for Lagrange multipliers, see some examples here: http://www.slimy.com/~steuard/teaching/tutorials/Lagrange.html#Examples
For your case, you minimize $f(x,y)$ subject to $g(x,y) = 0$, where $g(x,y) = x^2+y^2-1$. The multiplier $\lambda$ satisfies $\nabla f = \lambda \nabla g$, which for us is $(12x^2,6y) = \lambda (2x, 2y)$ and in addition we satisfy the constraint $g(x,y)=0$, so you have the system
12x^2 &= 2\lambda x\\
6y &= 2 \lambda y\\
x^2 + y^2 &= 1
From the second equation, clearly either $y=0$ (in which case $x = \pm 1$ from last equation) or $\lambda = 3$ (in which case $x=0$ from the first equation, and the last equation implies $y = \pm 1$). So the only solutions of the minimization problem are $(\pm 1, 0)$ and $(0, \pm 1)$. Evaluate to see which is better.
For the maximization problem, minimize $-f(x,y)$ instead.