Question

x^2 + √(x+y) + y^2 = 0; y is a differentiable function of x

Answer 1

that's a strange function. Since x^2 >=0 and y^2 >=0, you need x=y=0. Just one point.

Now, for x^2 + √(x+y) + y^2 = k for some other k>0,

2x + (1+y')/(2√(x+y)) + 2yy' = 0
y' = -(2x + 1/(2√(x+y))/(2y+1/(2√(x+y)))
= -(4x√(x+y)+1) / (4y√(x+y)+1)