To show that the first partial derivatives

$h_x\left(x\,y\right)\= \{\begin{array}{cc}\frac{y\left(x^4\+4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}& \left(x\,y\right)\ne \left(0\,0\right)\\ 0& \left(x\,y\right)\=\left(0\,0\right)\end{array}$

and

$h_y\left(x\,y\right)\= \{\begin{array}{cc}\frac{x\left(x^4-4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}& \left(x\,y\right)\ne \left(0\,0\right)\\ 0& \left(x\,y\right)\=\left(0\,0\right)\end{array}$

are continuous functions, first obtain the estimates

and$$

To show that $h_x$ is a continuous function, show that the limit of $\frac{y\left(x^4\+4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}$ is zero as $\left(x\,y\right)$ approaches the origin. To do this, show that the difference between this rational function and the purported limit of zero indeed goes to zero as $\left(x\,y\right)$ approaches $\left(0\,0\right)$. Using the first estimate from above, obtain

$\left|\frac{y\left(x^4\+4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}-0\right|$ $\le 2\left|y\right|$ 

which immediately implies that the limit is zero.

Similarly for $h_y$, the second estimate from above leads to

$\left|\frac{x\left(x^4-4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}-0\right|$ $\le 2\left|x\right|$ 

which again implies that the limit is zero.

Since, for $\left(x\,y\right)\ne \left(0\,0\right)$ both $h_x$ and $h_y$ are rational functions, Maple's bivariate limit command applies, and gives the following results immediately.

Since the limiting value and the declared value agree for each partial derivative, these first partials are continuous functions.