$$(1)\quad{y_1-f(x)\over x_1-x}={-1\over f'(x)}$$
Equation (1) is based on the fact that if $f(x)$ has some slope value 'm' at some point (x,y) on $f(x)$, then it also has a slope perpendicular to 'm' at the same point, and given by the negative reciprocal of 'm'.
If we let $f(x)=x^2$, then $f'(x)=2x$.
When $x_1=1$ , and $y_1=0$ , we have:
$${0-x^2\over 1-x}={-1\over 2x}$$
Clearing both sides of fractions and moving everything over to the left hand member we obtain:
$2x^3+x-1=0$
Solving for x we get: $x\approx0.589755...$
Since it's implied that $y=f(x)$, we can write $x=x_2\approx0.589755...$
and $f(x_2)=y_2\approx0.347810...$
We know that the point $(x_2,y_2)$ is on $f(x)$ because it satisfies the conditions of equation (1). We also know that the shortest distance from our known point $(x_1,y_1)$ not on $f(x)$, will lie on the normal line emanating from the point $(x_2,y_2)$. Consequently, the shortest distance is:
$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\approx0.537841...$