Verify if +-sqrt2 are roots of dy if 2y+y*x^2-2x=0 .

To verify if the given roots are the roots of dy = 0,
we'll have to differentiate the given function.

We'll try
to isolate y to the left side. For this reason, we'll add 2x both
sides:

2y+y*x^2 = 2x

Now,
we'll factorize by y:

y(x^2 + 2) =
2x

We'll divide by (x^2 + 2) both
sides:

y = 2x / (x^2 +
2)

Because the function is a ratio, we'll calculate it's
derivative using the quotient rule:

(u/v)'=
(u'*v-u*v')/v^2

dy =
[2x/(x^2+2)]'=[(2x)'*(2+x^2)-2x*(2+x^2)']/(2+x^2)^2

dy =
(2x^2+4-4x^2)/(2+x^2)^2

dy =
(4-2x^2)/(1+x^2)^2

We have, at numerator, a difference of
squares:

a^2-b^2=(a-b)(a+b)

(4-2x^2)
= (2-x*sqrt2)(2+x*sqrt2)

Because the denominator of dy  is
always positive, for any value of x, only the numerator could be
zero.

(2-x*sqrt2)(2+x*sqrt2)=0

We'll
set each factor as
zero.

2-x*sqrt2=0,

x*sqrt2=2

We'll
divide by sqrt2:

x = 2/sqrt2

x
= 2*sqrt2/2

x1 =
sqrt2

2+x*sqrt2 =
0

x*sqrt2 = -2

We'll divide by
sqrt2:

x =
-2/sqrt2

x2 =
-sqrt2

The roots of dy = 0 are
: {-sqrt2 ; sqrt2}.