The first partial derivatives of $g$, obtained in Example 4.11.1, are

$g_x\left(x\,y\right)\= \{\begin{array}{cc}2x\sin \left(\frac{1}{x^2\+y^2}\right)-\frac{2x\cos \left(\frac{1}{x^2\+y^2}\right)}{x^2\+y^2}& \left(x\,y\right)\ne \left(0\,0\right)\\ 0& \left(x\,y\right)\=\left(0\,0\right)\end{array}$

and

$g_y\left(x\,y\right)\= \{\begin{array}{cc}2y\sin \left(\frac{1}{x^2\+y^2}\right)-\frac{2y\cos \left(\frac{1}{x^2\+y^2}\right)}{x^2\+y^2}& \left(x\,y\right)\ne \left(0\,0\right)\\ 0& \left(x\,y\right)\=\left(0\,0\right)\end{array}$

To show that $g_x\left(x\,y\right)$ is not bounded, examine its limit as $\left(x\,y\right)\to \left(0\,0\right)$. The first term in $g_x$ is the product of $x$ with a factor that exhibits bounded oscillations. Since $x\to 0$, this term goes to zero in the limit. The second term is the product of a term that exhibits bounded oscillations multiplied by $x\/\left(x^2\+y^2\right)$. Since this factor, even on $y\=0$ is the unbounded $1\/x$, it should be clear that $g_x\left(x\,y\right)$ becomes unbounded as $\left(x\,y\right)\to \left(0\,0\right)$.

A similar analysis, mutatis mutandis, shows that $g_y\left(x\,y\right)$ becomes unbounded as $\left(x\,y\right)\to \left(0\,0\right)$.

Given that both $g_x$ and $g_y$ become unbounded at the origin, no further argument needs to be made that indeed, these first partials are discontinuous at the origin.