Show $\mathbb {R}[x,y]/(y^2-x, y-x)$ is not an integral domain

$\mathbb R [x,y] / (y-x) \cong \mathbb R[y]$ (trivially), and factoring by $(y^2-x)$ in the former means factoring by $(y^2-y)$ in the latter. And $\mathbb R[y]/(y^2-y)$ is not a domain, can you see why?

The golden rule. In a factor ring $R/I$ we have $a=0\bmod I$ iff $a\in I$ (where $a\in R$ ).

In your example $y^2=x\bmod I$ and $y=x\bmod I$. This gives us $y^2=y\bmod I$, so $y(y-1)=0\bmod I$, and this suggests that $y\bmod I$ and $y-1\bmod I$ are zero divisors in $R/I$. (Here $R=\mathbb R[x,y]$.)
However we have to check that they are not zero. For instance, suppose $y=0\bmod I$. Then, from the golden rule, $y\in I$, that is, $y\in (y^2-x,y-x)$. Now write $$y=(y^2-x)f(x,y)+(y-x)g(x,y),$$ and for $x=y=1$ we get $1=0$, a contradiction. (Do the same for $y-1\bmod I$.)

Hint: think of your ideal as $\langle y^2-x\rangle+\langle y-x\rangle$. So let us quotient $\mathbb R[x,y]$ by $\langle y-x\rangle$ first. What do we get? In this new ring, $y=x$, so plug that in to the other ideal to reduce our computation to a ring with only one variable! It should be much easier now.

Work through the above stuff and let me know if you want more hints