To check continuity of Multivariable functions To check continuity of function at origin given by 
$$F (x, y) = \begin{cases}\dfrac{xy^{2}}{x^{2} + y ^{4}}&;& \mbox{otherwise},\\
0&;&\mbox{ at origin}.
\end{cases}$$

 A: The inequality you wrote fails for $(x,y)=\left(\frac 1 3, \frac 35\right)$. To find counterexamples note the following.
For $x,y$ positive real numbers the inequality is equivalent to $xy^2\leq x^3+xy^4$ or $y^4-xy^2+x^2\ge 0$. 
Using the quadratic formula  on the LHS of the last inequality yields $y^2=\dfrac{1\pm \sqrt{1-4x^2}}{2}$ so now just pick $x$ such that $1-4x^2> 0$ (the strict inequality is imporant so you can get two zeroes of the $y$-equation $y^4-y^2x+x^2=0$). Letting $y_2$ be the largest zero and $y_1$ the smallest, choose $y^2\in ]\min(0, y_1), y_2[$ and you got yourself a counterexample.
Instead consider the sublimits $t\mapsto (kt^2,k)$ where $k\in \mathbb R$.
A: Consider the points ${\bf z}_n:=\bigl({1\over n^2},{1\over n}\bigr)$. Then $\lim_{n\to\infty}{\bf z}_n={\bf 0}$. But
$$\lim_{n\to\infty} f({\bf z}_n)=\lim_{n\to\infty}{{1\over n^2}\cdot\left({1\over n}\right)^2\over \left({1\over n^2}\right)^2+\left({1\over n}\right)^4}={1\over2}\ne f({\bf 0})\ .$$
Therefore $f$ is not continuous at ${\bf 0}=(0,0)$.
