Differential equation

y*(y'') - y' * (y^2) - (y'')^2 = 0

... it doesn't look hard but I just can't get it :| kinda blocked, can I get some help please

it is non-linear...does that help? lol do you'll end up with one of those weird graphs with saddle points etc...

Rewrite:
(y'')^2 -y(y'') +(y^2)y' = 0

Use quadratic formula:
y'' = [y ±√(y^2 -4(y^2)y'] /2

Simplify:
(2y''-y)^2 = y^2 (1 -4y')
Note: At this stage, it is clear that if y=0, then 2y''-y =0, so y''=0. This corresponds to the trivial
solution of y=0.

Simplify further:
(2y''/y^2 -1)^2 = 1 -4y'

At this stage, note that the equation holds when both sides equal the same constant.
(2y''/y^2 -1)^2 = 1 -4y' = C

You end up with two easier equations:
A) 1 -4y' = C
B) 2y''- (1±√C)*y^2 = 0

It kinda reads like you completed the square there Rick. You can skip the quadratic formula if you want.
(y''-1/2y)^2 = y''-y y''+1/4y^2
So the original is (y''-1/2y)^2 + (y'-1/4)y^2 = 0