Let $U$ be a uniform random variable on $[-1,1]$ and define the random variables $x_{K}=U^{K}$ for $K \in \mathbb{N}$ find the probability density function on $x_k$.
Answer the pdf is the derivative of the cdf which is $$F_{x_k}=P\bigg(U\in\bigg[-y^{\frac{1}{k}},y^{\frac{1}{k}}\bigg]\bigg)=y^{\frac{1}{k}}$$ so the pdf is $$f_{x_k}=\frac{1}{k}y^{\frac{1}{k}-1}$$ However the question gives a hint to look at $k$ odd and even separately so I'm assuming I've done something wrong.
Let it be that $F$ denotes the CDF of $U^k$ where $k$ is a positive integer.
1) $k$ is even.
Then $0\leq U^k\leq1$ a.s. so that $F(y)=0$ for $y<0$, $F(y)=1$ for $y\geq1$ and for $y\in(0,1)$:$$U^k\leq y\iff -y^{\frac1k}\leq U\leq y^{\frac1k}\text{ hence }F(x)=\frac12\cdot2y^{\frac1k}=y^{\frac1k}$$
Give this a sanity check for e.g. $k=2$.
2) $k$ is odd.
Then $-1\leq U^k\leq1$ a.s. so that $F(y)=0$ for $y<-1$, $F(y)=1$ for $y\geq1$ and for $y\in(0,1)$:$$U^k\leq y\iff U\leq y^{\frac1k}\text{ hence }F(x)=\frac12\cdot(y^{\frac1k}+1)$$
Give this a sanity check for e.g. $k=1$.
The PDF can be found easily as derivative of the CDF.