$U \sim \text{unif}(0, 1)$ and $3X^2 - 2X^3 - U = 0$. What is the p.d.f. of $X$? 
Let $U \sim \text{unif}(0, 1)$ and let $X$ be the root of the equation $3t^2 − 2t^3 − U = 0$. Show that $X$ has p.d.f.
$$f(x) = \begin{cases}6x(1-x) & 0\le x\le 1 \\ 0 & \text{otherwise} \end{cases}$$

Since $U \sim \text{unif}(0, 1)$, we have for $0\le a,b\le 1$, $P(a\le U\le b) = \frac{1}{b-a}$. We also have $3X^2 - 2X^3 - U = 0$. To find the p.d.f. of $X$, I first try to find the c.d.f. of $X$ (and differentiate it later). In fact one can check that $3x^2 - 2x^3$ is an increasing function in $0 \le x \le 1$. So,
$$P(X \le x) = P(3X^2 - 2X^3 \le 3x^2 - 2x^3)\\ = P(U \le 3x^2 - 2x^3) = \int_0^{3x^2 - 2x^3} \,dx = 3x^2 - 2x^3$$
Then
$$P(X \le x) = \begin{cases}3x^2 - 2x^3 & 0 \le x < 1 \\
1 &x \ge 1 \\0 &x < 0 \end{cases}$$
which gives the desired result on differentiation. I am wondering if there are any more direct approaches to this problem, i.e. without going through the c.d.f. if possible?

Here's a more direct approach (with extra words added in to explain it). Let $X$ be our unknown random variable with cdf $F_X(x)$. We know from Inverse Transform Sampling that the following relation holds
$$
X\overset{d}{=} F_X^{-1}(U),\quad U\sim\operatorname{Uniform(0,1)},
$$
where $\overset{d}{=}$ denotes equality in distribution. So it must be that $F_X(X)\overset{d}{=}U$ and thus, looking at your question $F_X(x)=3x^2-2x^3$. This answer, however, is not complete because we also need the support of $X$. Differentiating $F_X$ gives
$$
f(x)=6x(1-x),
$$
which is only nonnegative on the interval $x\in[0,1]$. Hence,
$$
F(x)=(3x^2-2x^3)1_{x\in[0,1]}+1_{x>1}.
$$
