#### Answer

Limit does not exist

#### Work Step by Step

We notice that if we directly substitute limits in the given function $f(x,y)=\frac{xy^{2}cosy}{x^{2}+y^{4}}$
Then $f(0,0)=\frac{0}{0}$
We get intermediate form. Therefore, we will calculate limit of function in following way.
To evaluate limit along x-axis; put $y=0$
$f(x,0)=\frac{xy^{2}cosy}{x^{2}+y^{4}}=\frac{x.0^{2}cos.0}{x^{2}+0^{4}}=0$
To evaluate limit along y-axis; put $x=0$
$f(0,y)=\frac{xy^{2}cosy}{x^{2}+y^{4}}=\frac{0,y^{2}cosy}{0^{2}+y^{4}}=0$
Approach (0,0) along another line let us say $x=y$ and $x\ne0$
$\lim\limits_{(x,y) \to (0,0)}f(y,y)=\lim\limits_{(x,y) \to (0,0)}f(y,y)\frac{y^{3}cosy}{y^{2}+x^{4}}=0$
Approach (0,0) along another line let us say $x=y^{2}$
$\lim\limits_{(x,y) \to (0,0)}f(y^{2},y)=\lim\limits_{(x,y) \to (0,0)}\frac{y^{2}.y^{2}cosy}{y^{4}+x^{4}}$
$=\lim\limits_{y \to 0}\frac{cosy}{2}$
$=\frac{cos0}{2}$
$=\frac{1}{2}$
For a limit to exist, all the paths must converge to the same point. Since, function $f(x,y)$ has two different values along two different paths, it follows that limit does not exist.
Hence, the limit does not exist.