Answer
$$y=\frac{1}{4}\sqrt{25-9x^2}$$
Work Step by Step
$y'=-\frac{9x}{16y}$
$16yy'=-9x$
$\int 16ydy=-\int 9xdx$
$8y^2=-9\frac{x^2}{2}+C$
$y^2=C-\frac{9x^2}{16}$
$y=\sqrt{C-\frac{9x^2}{16}}$
Now we need our equation to pass through the point $(1,1)$ so we apply it as an initial condition.
$1=\sqrt{C-\frac{9}{16}}$
$1=C-\frac{9}{16}$
$C=\frac{25}{16}$
So $y=\sqrt{\frac{25}{16}-\frac{9x^2}{16}}$
$$y=\frac{1}{4}\sqrt{25-9x^2}$$
