Let $G = \{1,x,y\}$ be a group with identity element $1$. Prove that $x^2 = y, y^2 = x,$ and $xy = 1.$ 
Let $G = \{1,x,y\}$ be a group with identity element $1$. Prove that $x^2 = y$, $y^2 = x$, and $xy = 1$.

Every time I try this I end up with pages of nonsense. Any help would be appreciated.

Let's fill in the Cayley table! Recall that Cayley tables are latin squares meaning that every element of the group appears exactly once in every row and every column (not counting the row and column headers). Let's start with a blank canvas:
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & & & \\ \hline \color{blue}x & & & \\ \hline \color{blue}y & & & \\ \hline\end{array}$$
First, we have that $1$ is the identity. Thus $1 \cdot 1 = 1$, $1 \cdot x = x \cdot 1 = x$, and $1 \cdot y = y \cdot 1 = y$, so
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & 1 & x & y \\ \hline \color{blue}x & x & & \\ \hline \color{blue}y & y & \color{red}{?} & \\ \hline\end{array}$$
What can we fill in for the red question mark$\color{red}?$ Note that $y$ already appears in the given row, so it's not $y$. Similarly, $x$ already appears in the given column, so it's not $x$. Thus, it must be $1$.
In more concrete terms, $y \cdot x$ cannot equal $y$, as
$$y \cdot x = y \implies y^{-1} \cdot y \cdot x = y^{-1} \cdot y \implies x = 1,$$
and similarly,
$$y \cdot x = x \implies y = 1.$$
Updating our Cayley table,
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & 1 & x & y \\ \hline \color{blue}x & x & & \color{red}{?} \\ \hline \color{blue}y & y & 1 & \\ \hline\end{array}$$
Similar logic works again here
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & 1 & x & y \\ \hline \color{blue}x & x & & 1 \\ \hline \color{blue}y & y & 1 & \\ \hline\end{array}$$
We can now finish the whole Cayley table by filling in the blanks. The second row/column is missing a $y$, and the third row/column is missing an $x$. Thus,
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & 1 & x & y \\ \hline \color{blue}x & x & y & 1 \\ \hline \color{blue}y & y & 1 & x \\ \hline\end{array}$$
We can now read off the fact that $x^2 = y$ and $y^2 = x$.
