The function $4x^3y/(x^4+y^2)$ fails the Lipschitz condition near the origin I have to prove that Lipschitz condition is  not satisfied for the function, 
$$ f(x) =
\begin{cases}
{4x^3y \over x^4 +y^2},  & \text{if $(x,y) \neq (0,0)$ } \\
0, & \text{if $(x,y)=(0,0)$ }
\end{cases}$$
throughout any domain which includes $(0,0)$.
I considered ,the domain $D = \{(x,y) : |x| \le a , |y|\le b,a \gt b\}$, and then considered,
$$f(x,y_1)-f(x,y_2) = 4x^3\left[{y_1 \over x^4+y_1^2} - {y_2 \over x^4+y_2^2}\right]$$
How to  proceed further?
 A: HINT: $f$ is locally Lipschitz near the origin if and only if its partial derivatives are bounded near the origin. 
For $(x,y) \ne (0,0)$ we have 
\begin{eqnarray*}
\frac{\partial f}{\partial x}(x,y) &=& -\frac{4 x^2 y (x^4 - 3 y^2)}{(x^4 + 
  y^2)^2}\\
\frac{\partial f}{\partial y}(x,y) &=&\frac{4 x^3 (x^2 - y) (x^2 + y)}{(x^4 + y^2)^2}
\end{eqnarray*}
These expressions do not look much nicer but notice that 
\begin{eqnarray*}
\frac{\partial f}{\partial y}(x,0) &=& \frac{4}{x}
\end{eqnarray*}
A: What Lipschitz continuity implies for a function is that for you could find a constant $K$ such that for any pair of points ${X_1}$ and ${X_2}$ in the domain under study, the distance between the images of these points under the function is less than $K$ times of the distance between the original points. To prove that a function is not Lipschitz continuous, we could prove that a quotient defined as the distance of the images divided by the distance of the points diverges at least for some points in the domain. Ok, these said, assume your first point to be $(0,0)$, which by your definition the image of it under your function is $0$. Now, the quotient reads as $$\frac{{d(f({x_2},{y_2}),f({x_1},{y_1}))}}{{d(({x_2},{y_2}),({x_1},{y_1}))}} = \frac{{d(f({x_2},{y_2}),0)}}{{d(({x_2},{y_2}),(0,0))}}$$where $d$ is a suitably defined measure of the distance. Assuming the Euclidean norm, we have $$\frac{{d(f({x_2},{y_2}),0)}}{{d(({x_2},{y_2}),(0,0))}} = \frac{{\sqrt {{{\left( {\frac{{4{x^3}y}}{{{x^4} + {y^2}}}} \right)}^2} - {0^2}} }}{{\sqrt {{{(x - 0)}^2} + {{(y - 0)}^2}} }} = \frac{{\left| {\frac{{4{x^3}y}}{{{x^4} + {y^2}}}} \right|}}{{\sqrt {{x^2} + {y^2}} }}$$Now, to simplify the matters, switch to the polar coordinates defined as $x = r\cos \theta$ and $y = r\sin \theta$ to get $$\frac{{d(f({x_2},{y_2}),0)}}{{d(({x_2},{y_2}),(0,0))}} = \frac{{\left| {\frac{{4{r^4}{{\cos }^3}\theta \sin \theta }}{{{r^4}{{\cos }^4}\theta  + {r^2}{{\sin }^2}\theta }}} \right|}}{r} = \left| {\frac{{4r{{\cos }^3}\theta \sin \theta }}{{{r^2}{{\cos }^4}\theta  + {{\sin }^2}\theta }}} \right|$$now, you should prove that this quotient does not exist.
