$R_P\cong \mathbb{R}[x]_{(x-1)}$

A bad habit (especially when one works with quotient rings of polynomial rings) is to denote the indeterminates by small letters. So, let us write $R=K[X,Y]/(XY)$ as it should be. (Here $K$ is an arbitrary field.) Denote the residue classes of $X$ and $Y$ by $x$ and $y$, and thus $R=K[x,y]$. In this ring one considers the maximal ideal $P=(x-1)$. (In order to show that $P$ is maximal we consider $R/P=K[x,y]/(x-1)=\frac{K[X,Y]/(XY)}{(X-1,XY)/(XY)}$ which is isomorphic to $K[X,Y]/(X-1,XY)$. It is not hard to show that the later is isomorphic to $K$.)

Now come back to the main question: in fact, $R_P=K[x,y]_{(x-1)}$ is $(K[X,Y]/(XY))_{(X-1,XY)/(XY)}$, which at its turn is isomorphic to $K[X,Y]_{(X-1,XY)}/(XY)K[X,Y]_{(X-1,XY)}$. But $(XY)K[X,Y]_{(X-1,XY)}=(Y)K[X,Y]_{(X-1,XY)}$ since $X\notin(X-1,XY)$. Therefore we get $R_P\simeq K[X,Y]_{(X-1,XY)}/(Y)K[X,Y]_{(X-1,XY)}\simeq K[X]_{(X-1)}$.