Find the minimum value of \frac{x^2}{x - 1 + x^3} for x > 1
We start by getting a common denominator, to rewrite the denominator of the expression $\frac{x^2}{x-1+x^3}$. Combining like terms, we get \[
\begin{aligned} x-1 + x^3 &= -1+x+x^3 \\&= -1+x^3+x. \end{aligned}
\]Using the Simon's Favorite Factoring Trick, we add and subtract 1 to obtain \[
\begin{aligned} -1+x^3+x &= -1+1+x+x^3 \\ &= (x^3+x) + 1 + (-1) \\ &= x^2(x) + x^2(1/x) +1 +(-1). \end{aligned}
\]Using this factorization, we rewrite the original expression as \[
\frac{x^2}{-1+x^3+x} = \frac{x^2(x)}{-1+x^3+x} = \frac{x^3}{x^3 + x + (-1)}.
\]We can then rewrite the fraction as \[
\frac{x^3}{x^3 + x + (-1)} = \frac{x^3 + x + (-1) - (x+(-1))}{x^3 + x + (-1)} = \frac{x^3 + x + (-1)}{x^3 + x + (-1)} - \frac{x+(-1)}{x^3 + x + (-1)}.
\]Simplifying, we get \[
\frac{x^3 + x + (-1)}{x^3 + x + (-1)} - \frac{x+(-1)}{x^3 + x + (-1)} = 1 - \frac{x+(-1)}{x^3 + x + (-1)}.
\]Notice that the denominator of $1$ is positive for $x >1 ,$ and we know that $x + (-1)>0$ since $x > 1$. Hence $1 - \frac{x+(-1)}{x^3 + x + (-1)} > 1-1 = 0$ for $x > 1$. This means that for all $x>1$, $\frac{x^2}{x-1+x^3} = 1 - \frac{x+(-1)}{x^3 + x + (-1)} > 0$.
We can actually compute the value of this expression as $\boxed{0}$. (Notice that as $x\to 1^+$, the expression grows arbitrarily large.)