The point on the curve $y=x^2$  which is nearest to (3,0) is

$\begin{array}{rl} & PQ=\sqrt{(x-3{)}^2+(y-0{)}^2}\\ & \text{Let}z=(x-3{)}^2+y^2\\ & z=x^2-6x+9+x^4\\ & \frac{dz}{dx}=2x-6+4x^3\end{array}$
$\begin{array}{rl} & \mathrm{Z}\text{is minimum}\frac{dz}{dx}=0\\ & 2x-6+4x^3=0\\ & x=1\\ & \text{If}x=1\text{then}y=x^2\\ & \Rightarrow y=1\\ & \Rightarrow (x,y)=(1,1)\end{array}$