Continuity of $f(x,y)=4x^3y^{11}(x^4+y^8)^{-2}$ at $(0,0)$

The limit at $(0,0)$ exists and is $0$.

To prove this, note that, for every $\theta$ in $(0,1)$ and every nonnegative $(u,v)$, $$u^\theta\cdot v^{1-\theta}\leqslant\max\{u,v\}^\theta\cdot\max\{u,v\}^{1-\theta}=\max\{u,v\}\leqslant u+v.$$ Thus, for every $(x,y)$, $$x^4+y^8\geqslant |x|^{4\theta}\cdot|y|^{8(1-\theta)},$$ which implies that $$ \left|\frac{x^3y^{11}}{(x^4+y^8)^2}\right|\leqslant |x|^{3-8\theta}\cdot|y|^{16\theta-5}. $$ The RHS goes to $0$ when $(x,y)$ goes to $(0,0)$ as soon as both exponents $3-8\theta$ and $16\theta-5$ are nonnegative and at least one of them is positive, that is, for every $\theta$ such that $\theta\geqslant\frac5{16}$ and $\theta\leqslant\frac38$. Since $\frac5{16}\leqslant\frac38$, this interval is not empty, which proves the result.

For example, $\theta=\frac13$ yields $$ \left|\frac{x^3y^{11}}{(x^4+y^8)^2}\right|\leqslant |x|^{1/3}\cdot|y|^{1/3}\to0. $$ More generally, for every positive $(a,b,c,d,e)$, $$ \frac{x^ay^b}{(x^c+y^d)^e}, $$ goes to $0$ at $(0,0)$ as soon as $$ \frac{a}c+\frac{b}d> e. $$ Note that $$ \frac{3}4+\frac{11}8=\frac{17}8>2. $$

For all $(x,y)$ we have $$(x^4+y^8)^2\ge y^{16}\quad\hbox{and}\quad (x^4+y^8)^2\ge2x^4y^8\ .$$ Now suppose that $(x,y)\ne(0,0)\,$. If $|x|\le y^2$ we have $y\ne0$ and $$|f(x,y)|\le\frac{4|x|^3|y|^{11}}{|y|^{16}}\le4|y|\ ,$$ while if $|x|\ge y^2$ we have either $y=0$, when $f(x,y)=0$, or $y\ne0$ and $x\ne0$ and $$|f(x,y)|\le\frac{4|x|^3|y|^{11}}{2|x|^4|y|^8}=\frac{2|y|^3}{|x|}\le2|y|\ .$$ Therefore for all $(x,y)\ne(0,0)$ we have $$|f(x,y)|\le4|y|\ ,$$ and the right hand side tends to zero as $(x,y)\to(0,0)$.

Cauchy inequality (arithmetic means larger or equal to geometric mean) implies that for every $a,b\ge0$ $$ \underbrace{\frac{a}{k}+\cdots+\frac{a}{k}}_{k\,\,\text{times}} +\underbrace{\frac{b}{\ell}+\cdots+\frac{b}{\ell}}_{\ell\,\,\,\text{times}} \ge (k+\ell)\,\,\sqrt[k+\ell]{\left(\frac{a}{k}\right)^k\left(\frac{b}{\ell}\right)^\ell} $$ or $$ a+b\ge c_{k,\ell}a^{\frac{k}{k+\ell}}b^{\frac{\ell}{k+\ell}} $$ for a suitable constant $c_{k,\ell}>0$.

So in our case ($a=x^4$ and $b=y^8$) $$ x^4+y^8\ge c_{3,5}(x^4)^{3/8}(y^8)^{5/8}=c_{3,5}\lvert x\rvert^{3/2}\lvert y\rvert^5, $$ or $$ (x^4+y^8)^2\ge c_{3,5}^2\lvert x\rvert^3 \lvert y\rvert^{10} $$ and hence $$ \lvert f(x,y)\rvert=\frac{4\lvert x^3y^{11}\rvert}{(x^4+y^8)^2}\le \frac{\lvert y\rvert}{c_{3,5}^2}. $$ Clearly $$ \lim_{(x,y)\to(0,0)}f(x,y)=0. $$