Let $a$ be any real number.
$\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }x=a^2$ (when $x\to a$ through rational values)
$\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }\left(2-x\right)=2-a$ (when $x\to a$ through irrational values)
$\underset{x\to a}{\lim }f\left(x\right)$ will exist only when $a^2=2-a$
i.e., when $a=1,\mathrm{}-2$
Thus, if $a\ne 1$ or $-2\mathrm{}$ then $\underset{x\to a}{\lim }f\left(x\right)$ will not exist
Hence, $f\left(x\right)$ will be discontinuous at $x=a$ where $a\ne 1$ or $-2$