1. ## pathological function question

let f : R^2 -> R be defined as:
f = x^3 / (x^2 + y^2) if (x,y) =/= 0
and 0 if (x,y) = 0.

let g be any curve in R^2 that passes through the origin. then show that f o g is differentiable while f itself is not. so i have that g: R -> R^2 and using the definition of the limit i have lim (h->0) [f(g(h)) - f(g(0)) - mh]/h which we want to show should = 0. i am evaluating the derivative at 0 because at any other point where (x,y) =/= 0 then f is differentiable so the only trouble spot comes from (x,y) = 0. however i am stuck since i don't know anything about g, just that its a curve that passes through the origin which does not help me figure out what f(g(h)) could be. any help is greatly appreciated. thanks.

2. The "trouble spot" for f is the origin, so for $f\circ g$ it would be when g passes through the origin. Using the chain rule, what would be the expression for the derivative of $f\circ g$ at the trouble spot?

3. so i'm assuming g: R -> R^2 is given by g(t) = (u(t), v(t)) so D(fog) = f_u * u'(t) + f_v * v'(t). so at 0 it would be m = D(fog(0)) = f_u(0) * u'(0) + f_v(0) * v'(0). i am not quite sure what to do next. i have the partial of f with respect to u which i don't know how to evaluate. trying to use the limit definition, f(g(h)) would become u(h)^3 / ((u(h)^2 + v(h)^2) which doesn't look promising. i know that at least one of u(t) or v(t) is not constant because if they both were, it would just be a point. so the quotient should be some function of h but there is no telling what kind of function of h it is and whether or not it will allow the limit to be 0. is the limit definition the way to go with this problem?

4. does anyone have ideas on how to tackle this further?