Maple determines the bivariate limit at the origin to be 1.

To see why the bivariate limit at the origin might have the value 1, divide the  numerator and denominator of $f$ to put $f\left(x\,y\right)$ into the form

$\frac{1\+\frac{x^4\+x y^2}{x^2\+y^2}}{1-\frac{x^{2}y}{x^2\+y^2}}$ 

The rational functions in the numerator and denominator of this form of $f$ both tend to zero as $\left(x\,y\right)\to \left(0\,0\right)$. Indeed, Maple provides the following bivariate limits, obtained either through the Context Panel, or by application of the limit command with the appropriate syntax.

Clearly, then, the bivariate limit of $f$ as $\left(x\,y\right)\to \left(0\,0\right)$ will necessarily be 1, in which case the required extension is

$g\left(x\,y\right)\=\{\begin{array}{cc}f\left(x\,y\right)& \left(x\,y\right)\ne \left(0\,0\right)\\ 1& \left(x\,y\right)\=\left(0\,0\right)\end{array}$

Analytic justification for Auxiliary Limit (1) is based on the following estimate.

Consequently, $\left|\frac{x^4\+x y^2}{x^2\+y^2}\right|\le x^2\+y^2\+\sqrt{x^2\+y^2}$, and the limiting value of zero is established.

Analytic justification for Auxiliary Limit (2) is based on the following estimate.

Consequently, $\left|\frac{x^{2}y}{x^2\+y^2}\right|\le \sqrt{x^2\+y^2}$, and the limiting value of zero is established.

Indeed, $\mathrm{limit}\left(f\left(x\,y\right)\,\left\{x\=0\,y\=0\right\}\right)$ = $1$.