The first partial derivatives of $g$, obtained in Example 4.11.1, are
${g}_{x}\left(x\,y\right)\=lcub;\begin{array}{cc}2x\mathrm{sin}\left(\frac{1}{{x}^{2}plus;{y}^{2}}\right)-\frac{2x\mathrm{cos}\left(\frac{1}{{x}^{2}plus;{y}^{2}}\right)}{{x}^{2}plus;{y}^{2}}& \left(xcomma;y\right)\ne \left(0comma;0\right)\\ 0& \left(xcomma;y\right)equals;\left(0comma;0\right)\end{array}$
and
${g}_{y}\left(x\,y\right)\=lcub;\begin{array}{cc}2y\mathrm{sin}\left(\frac{1}{{x}^{2}plus;{y}^{2}}\right)-\frac{2y\mathrm{cos}\left(\frac{1}{{x}^{2}plus;{y}^{2}}\right)}{{x}^{2}plus;{y}^{2}}& \left(xcomma;y\right)\ne \left(0comma;0\right)\\ 0& \left(xcomma;y\right)equals;\left(0comma;0\right)\end{array}$
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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 $\left(x\,y\right)\to \left(0\,0\right)$.
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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.