Answer
$$y=\sqrt{1-\left(\sqrt{1-x^2}-1\right)^2}$$
Work Step by Step
$y\sqrt{1-x^2}y'-x\sqrt{1-y^2}=0$
$y\sqrt{1-x^2}y'=x\sqrt{1-y^2}$
$\frac{yy'}{\sqrt{1-y^2}}=\frac{x}{\sqrt{1-x^2}}$
$\int \frac{y}{\sqrt{1-y^2}}dy=\int \frac{x}{\sqrt{1-x^2}}dx$
$-\sqrt{1-y^2}=-\sqrt{1-x^2}+C$
$\sqrt{1-y^2}=\sqrt{1-x^2}+C$
$1-y^2=\left(\sqrt{1-x^2}+C\right)^2$
Let's apply our initial condition here.
$1-1^2=(\sqrt{1}+C)^2$
$0=(1+C)^2$
$C=-1$
Solving for $y$ and plugging in $C=-1$ we end up with
$$y=\sqrt{1-\left(\sqrt{1-x^2}-1\right)^2}$$
