The shortest distance between the line $y-x=1$ and the curve $x=y^2$ is:

Curve: $y^2=x$. Line: $x-y+1=0$.Shortest distance occurs when the tangent to the curve is parallel to the line.Slope of line $=1$.Derivative of curve: $2y{y}^{\prime }=1⟹y^{\prime }=1/(2y)$.Set $1/(2y)=1⟹y=1/2$.Then $x=y^2=1/4$.Point $P(1/4,1/2)$.Distance $=\frac{|1/4-1/2+1|}{\sqrt{1^2+(-1{)}^2}}=\frac{|3/4|}{\sqrt{2}}=\frac{3}{4\sqrt{2}}=\frac{3\sqrt{2}}{8}$.