Can anyone solve the differential equation of x(1+y^2)^1/2dx=y(1+x^2)^1/2dy?

The answer is as follows:

Looks separable to me

x * dx / (1 + x^2)^(1/2) = y * dy / (1 + y^2)^(1/2)

u = (1 + x^2)^(1/2)
u^2 = 1 + x^2
2u * du = 2x * dx
u * du = x * dx

u * du / u =>
du

v = (1 + y^2)^(1/2)
v^2 = 1 + y^2
2v * dv = 2y * dy
v * dv = y * dy

y * dy / (1 + y^2)^(1/2)
v * dv / v =>
dv

Now we have:

du = dv

Integrate

u + C = v
(1 + x^2)^(1/2) + C = (1 + y^2)^(1/2)

You can expand from there if you want.

Hint: From highschool algebra, do you see that this is separable? 
Separate it first.  What do you get? Answer that & we will proceed. 

Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them. Too late!

x√( 1 + y² ) dx = y√( 1 + x² ) dy
x / √( 1 + x² ) dx = y / √( 1 + y² ) dy
∫x / √( 1 + x² ) dx = ∫ y / √( 1 + y² ) dy

LHS:
let u = ( 1 + x² )  such that 
du/dx = 2x
dx = du/(2x)
∴ 
∫ x / √( 1 + x² ) dx 
= ∫ ( x /√u )  *  du/(2x)
= ½ ∫ u^(-½) du
= √u
= √( 1 + x² ) + constant 


RHS:
∴ 
∫ y / √( 1 + y² ) dy 
= √( 1 + y² ) + constant 

THUS:
√( 1 + y² ) = √( 1 + x² ) + C 
1 + y²  = [ √( 1 + x² ) + C ]²
y² =  [ √( 1 + x² ) + C ]² - 1
y² = C² + x² + 2C√( x² + 1 )

y = ±√[ C² + x² + 2C√( x² + 1 ) ]