Handling dy/dx on both sides of equations: dy/dx = 0.5*y*y/(dy/dx) = 0.5*y^2/(dy/dx)

[Tigers!]

Gudday all
I have been revisiting calculus many years after university and try to learn again.
Been doing some investigations with sub-tangents and sub-normals and getting stuck with handling multiple dy/dx.
A. Let dy/dx at point P on a curve = area of triangle = 0.5*bh where h =y and b= sub-tangent at P
i.e. dy/dx = 0.5*y*y/(dy/dx)
= 0.5*y^2/(dy/dx)

How is the dy/dx on the RH side handled?
Do w e just multiply both sides by dy/dx to get d^2y/dx^2 = 0.5*y^2
and then just double integrate or something else?

B. Let dy/dx at point P on a curve = area of triangle = 0.5*bh where h =y and b= sub-normal at P
i.e. dy/dx = 0.5*y*y*dy/dx
= 0.5*y^2*dy/dx
Does this become 0.5*y^2*dy/dx - dy/dx = 0
then dy/dx(0.5*y^2 - 1) = 0
Separating the variables to get [integral sign] (0.5*y^2-1) dy = [integral sign] 0 dx
Then integrate or is this up the creek?

 

[tkhunny]

I'm getting no sense at all of what it is you are trying to do. dy/dx on both sides may make very little sense.
Can you provide a single example and demonstrate your best efforts?

 

[Tigers!]

Sorry. Hopefully this will be better.

 

[tkhunny]

You already said that. How did you get dy/dx in the denominator, anyway. Typically, if you are drawing a pretty picture, like that, dy and dx would appear separately. Thus, your confusion goes away.

The subtraction version is unlikely ever to lead you anywhere.