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

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

and

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

The following estimate shows that $f_x\left(x\,y\right)$ is bounded.

where the inequality $\left|x\right|\le \sqrt{x^2\+y^2}$ is Inequality 4 in Table 3.2.1.

The following estimate shows that $f_y\left(x\,y\right)$ is bounded.

where the inequality $\left|y\right|\le \sqrt{x^2\+y^2}$ is Inequality 5 in Table 3.2.1.

To show that $f_x\left(x\,y\right)$ is discontinuous at the origin, consider its limit as $x\to 0$ along the $x$-axis. Since

the limit of the first term is zero, but the limit of the second term does not exist because of the infinite oscillations in the cosine term. Hence, $\underset{x\to 0}{lim}{f}_x\left(x\,0\right)$ does not exist, so the bivariate limit as $\left(x\,y\right)\to \left(0\,0\right)$ cannot exist. Hence, $f_x\left(x\,y\right)$ is not continuous at the origin.

To show that $f_y\left(x\,y\right)$ is discontinuous at the origin, consider its limit as $y\to 0$ along the $y$-axis. Since

the limit of the first term is zero, but the limit of the second term does not exist because of the infinite oscillations in the cosine term. Hence, $\underset{y\to 0}{lim}{f}_y\left(0\,y\right)$ does not exist, so the bivariate limit as $\left(x\,y\right)\to \left(0\,0\right)$ cannot exist. Hence, $f_y\left(x\,y\right)$ is not continuous at the origin.