Answer
$\left(-\frac{1}{3},±\frac{4\sqrt 2}{3}\right)$
Work Step by Step
From the figure, we see that there are two points that are farthest away from $A(0,1)$.
The distance $d$ from $A$ to an arbitrary point $P(x,y)$ on the ellipse is
$d$ = $\sqrt {(x-1)^{2}+(y-0)^{2}}$
and the square of the distance is
$S$ = $d^{2}$ = $x^{2}-2x+1+y^{2}$ = $x^{2}-2x+1+(4-4x^{2})$ = $-3x^{2}-2x+5$
$S'$ = $-6x-2$
$S'$ = $0$
$x$ = $-\frac{1}{3}$
$S''$ = $-6$ $\lt$ $0$ so we know that $S$ has a maximum at $x$ = $-\frac{1}{3}$.
Since $-1$ $\leq$ $x$ $\leq$ $1$
$S(-1)$ = $4$
$S\left(-\frac{1}{3}\right)$ = $\frac{16}{3}$
$S(1)$ = $0$
We see that the maximum distance is $\sqrt {\frac{16}{3}}$.
The corresponding $y$-values are
$y$ = $±\sqrt {4-4\left(-\frac{1}{3}\right)^{2}}$ = $±\frac{4\sqrt 2}{3}$
The points are $\left(-\frac{1}{3},±\frac{4\sqrt 2}{3}\right)$.
