How do i find the second derivative of x^2 + xy = 5

2x+y+x dy/dx = 0
dy/dx = -(2x+y)/x = -y/x - 2
d^2y/dx^2 = [-x dy/dx + y]/x^2
d^2y/dx^2 = [-x(-y/x - 2) + y]/x^2
d^2y/dx^2 = (2y+2x)/x^2 = 2(y+x)/x^2

The first derivative of this function; 2x+yy'=0 then second der of it ; 2+y^2(y'')=0 that brings y''=(-2)/y^2

x^2 + xy = 5
2x + y + xy' = 0
2 + y' + y' + xy" = 0
xy" = -2-2y'
y" = -2(1+y')/x