Differentiability of f o g at the origin

[demonelite123]

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.

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?

any help is greatly appreciated. thanks.

 

[demonelite123]

i am still having trouble with this problem. the only way i know of to show a function is differentiable is to use the limit definition which is lim (h->0) 1/h (f(x+h)-f(x)-Df(x)(h)) = 0. when checking if f(g(t)) satisfies this requirement, f(g(0)) is 0 obviously but i have trouble showing that f(g(h)) = Df(g(0))(h). because g is arbitrary, i don't see a way i can make sense of the derivative of it since it can be many different things. how would one attack this problem? am i missing something important?