determine whether $f(x, y) = \frac{xy^3}{x^2 + y^4}$ is differentiable at $(0, 0)$. I am new to multivariable calculus and my textbook doesn't give out solutions so I'm just wondering how you go about proving something like this? I know that a function is differential at a point $a$ if it's continous at $a$ and and the partial derivatives of $f$ exist near $a$ but I have never actually seen an example. Heres the question:  
Assume $f(0, 0) = 0$, and determine whether $f(x, y) = \frac{xy^3}{x^2 + y^4}$ is differentiable at $(0, 0)$.
 A: Compute the partial derivative:
$${f_x'} (0,0)=\mathop {\lim }\limits_{x \to 0} {{f(x,0) - f(0,0)} \over x} = \mathop {\lim }\limits_{x \to 0} {{0 - 0} \over x} = 0$$
$${f_y'} (0,0)=\mathop {\lim }\limits_{y \to 0} {{f(0,y) - f(0,0)} \over y} = \mathop {\lim }\limits_{x \to 0} {{0 - 0} \over y} = 0$$
Compute the limit:
$$\mathop {\lim }\limits_{ x \to 0 \atop y \to 0} \frac {f(x,y) - {f_x'} (0,0)x - {f_y'} (0,0)y} {\sqrt {{x^2} + {y^2}} } $$
$$= \mathop {\lim }\limits_{ x \to 0 \atop y \to 0} \frac {xy^3} {(x^2+y^4)\sqrt {{x^2} + {y^2}} }$$
$$\xrightarrow{x=ky^2} \mathop {\lim }\limits_{y \to 0} {\frac{k{y^5}} {({k^2} + 1){y^4}\sqrt {{k^2}{y^4} + {y^2}} }}$$
$$=\mathop {\lim }\limits_{y \to 0} {\frac{k} {({k^2} + 1)\sqrt {k^2 y^2 + 1} }}=\frac{k}{k^2+1}$$
If $f(x,y) - {f_x'} (0,0)x - {f_y'(0,0)x} = o(\sqrt {{x^2} + {y^2}}) $ , which is equivalent to $ \mathop {\lim }\limits_{ x \to 0 \atop y \to 0} \cfrac {xy^3} {(x^2+y^4)\sqrt {{x^2} + {y^2}}}=0$ , we can say $f(x,y)$ is differentiable at $(0,0)$ , but in this case , the limit is relating to $k$ so that the limit doesn't exist, that is to say $f(x,y)$ is non-differential at $(0,0)$.
A: Check the partial derivatives, if they are continuous at $(0,0)$. If they are then, $f$ is differentiable at $(0,0)$.
A: Using the following claim: 

Partial derivatives exist and they are continuous at $(0,0)$, then it is  differentiable at  $(0,0)$.

