Verified answer

dy/dx = -x/y

dy = dx * -x/y

y dy = -x dx

1/2 y^2 = -1/2 x^2 + constant

y^2 = -x^2 + c

y = √(c - x^2) or y = -√(c - x^2)

Perhaps you can just recognize that if

x^2 + y^2 = r^2

2x + 2ydy/dx = 0

dy/dx = -x/y

It is the slope at a given point (x, y) on a circle.

Or in reverse

ydy = -xdx

(1/2)y^2 =( -1/2)x^2 + C

x^2 + y^2 = 2C which it is convenient to rename as r^2

Regards - Ian H

dy/dx = -x/y, separating out the variables,

y.dy=-x.dx, integrating,

y²/2 = -x²/2 + K {arbitrary constant}, simplifying

y² + x² = M {M=2K}

y*dy = -x*dx

Too easy.