Use implicit differentiation to find dy/dx

xy + x + y = x^2y^2

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

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

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

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

For problems like this, just multiply both sides by d/dx

d/dx(xy+x+y)=d/dx(x^2y^2)

x*y'+y'+y+1=2x^2y*y'+2xy^2

Just factor and divide
y'(x) = (-2 x y^2+y+1)/(2 x^2 y-x-1)

In general, if you have to do differentiation, you can use Wolfram Alpha, its software is good enough to
take in requests like "xy + x + y = x^2y^2, dy/dx=".
it will give you the answer, and show you step by step how it got there using the basic rules of calculus

[1∙y + x∙(dy/dx)] + 1 + (dy/dx) = [2x∙y² + x²∙(2y)(dy/dx)]

y + x(dy/dx) + 1 + (dy/dx) = 2xy² + 2x²y(dy/dx)

(dy/dx)[x + 1 - 2x²y] = 2xy² - y - 1

dy/dx = [2xy² - y - 1] / [1 + x - 2x²y]

Or you could do:
g(x,y) = h(x,y)
f(x,y) = h(x,y) - g(x,y)

dy/dx = -- fx(x,y) / fy(x,y)

Differentiating implicitly:

xy' + y + 1 + y' = 2(xy)(xy' + y)

solve for y'.