Find the shortest distance from the point $P(0,1)$ to a point on the curve $x² - y² = 1$ , and find the point on the curve closest to $P$. 
Find the shortest distance from the point $P(0,1)$ to a point on the curve $x² - y² = 1$, and find the point on the curve closest to $P$.

What I did so far is :
plot the y var from $x^2-y^2=1 \implies y=\sqrt{1-x^2}$
Create a distance equation : $d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
$ d = \sqrt{x^2+(1-y)^2}$
$ d = \sqrt{2-2\sqrt{1-x^2}}$
$\displaystyle \frac d{dx} (d) =\frac x{\sqrt{1-x^2}\sqrt{2-2\sqrt{1-x^2}}}$
I need to find the max or min?
Any suggestions? thanks!

Your approach is fine.  It is a little easier to minimize $d^2$ than $d$ as that gets rid of one square root.  You lost a sign when you went from $d=\sqrt{x^2+(1-y)^2}$ to $d=\sqrt{2-2\sqrt{1-x^2}}$ as the $y^2$ term is positive.  Plotting the curve shows that the distance to either branch from $(0,1)$ is the same, so we can use the positive square root $x=\sqrt{1+y^2}$.  We have $$d^2=x^2+(1-y)^2=2-2y+2y^2\\\frac {d(d^2)}{dy}=-2+4y$$ which we set to zero and find $y=\frac 12, x=\sqrt{\frac 54}$.  The squared distance is then $d^2=\frac 54+\frac 14=\frac 32$ and the linear distance is $d=\sqrt \frac 32$.  This must be a minimum, as there are points on the hyperbola very far away.
