Implicit function on functions composed of itself

[chy1013m1]

Suppose F(x, y) is C1. F(0, 0) = 0. What conditions on F will guarantee that the equation F(F(x, y), y) = 0 can be solved for y as a C1 function of x near (0, 0) ?

would it simply be dF/dy not equal 0 ?

 

[Tom1992]

Use the implicit function theorem. Seems you are on the right track, but F is a composition of F and y, so you would have to use the chain rule to right out dF/dy properly.