Determine whether $f(x,y)=\sqrt{|xy|}$ and/or $g(x,y)=e^{|x|^3y}$ are differentiable at the point $(0, 0)$. Determine whether $$f(x,y)=\sqrt{|xy|}$$ and/or $$g(x,y)=e^{|x|^3y}$$ are differentiable at the point $(0, 0)$. 
Also, find its total derivative if it exists at (0, 0); if not, prove that it is not differentiable at $(0, 0)$.
My approach:
For $f(x,y)$, I got it is not differentiable because the limit tends to $1$ and $-1$. But I am not sure about $g(x,y)$.
 A: For $g$, the exponential does not change the differentiability. So we can study first $(x,y) \mapsto |x|^3 y$, which is differentiable with total derivative 0.
A: I would solve the first one. Follow the same to solve the second one as well.
$${f_x} (0,0)=\mathop {\lim }\limits_{x \to 0} {{f(x,0) - f(0,0)} \over x} = \mathop {\lim }\limits_{x \to 0} {{0 - 0} \over x} = 0$$
and 
$${f_y} (0,0)=\mathop {\lim }\limits_{y \to 0} {{f(0,y) - f(0,0)} \over y} = \mathop {\lim }\limits_{x \to 0} {{0 - 0} \over y} = 0.$$
Therefore
$$I=\lim_{(x,y)\rightarrow (0,0) } {{f(x,y) - f_x(0,0)x-f_y(0,0)y-f(0,0)} \over {\sqrt{x^2+y^2}}} = \lim_{(x,y)\rightarrow (0,0) } {\sqrt{|xy|}\over {\sqrt{x^2+y^2}}}.$$
If the limits are taken along the path $y=mx$ then $I= \sqrt{|m|}/\sqrt{1+m^2} =g(m)\neq 0.$
Therefore $f$ is not differentiable at the origin.
[Note for $f$ to be differentiable at $(0,0), $ I mut be $0 $ at that point ].
For the second;
$${g_x} (0,0)=\mathop {\lim }\limits_{x \to 0} {{g(x,0) - g(0,0)} \over x} = \mathop {\lim }\limits_{x \to 0} {{1 - 1} \over x} = 0$$
$${g_y} (0,0)=\mathop {\lim }\limits_{y \to 0} {{g(0,y) - g(0,0)} \over y} = \mathop {\lim }\limits_{x \to 0} {{1 - 1} \over y} = 0.$$
Therefore
$$J=\lim_{(x,y)\rightarrow (0,0) } {{g(x,y) - g_x(0,0)x-g_y(0,0)y-g(0,0)} \over {\sqrt{x^2+y^2}}} = \lim_{(x,y)\rightarrow (0,0) } {{e^{|x|^3y}-0-0-1}\over {\sqrt{x^2+y^2}}}$$
$$=\lim_{(x,y)\rightarrow (0,0) } {({e^{|x|^3y}-1)}\over {|x|^3y}}{|x|^3y\over{\sqrt{x^2+y^2}}}=0, $$ because
the first term has limit value 1 and the limit value for the second term  is $0.$ Calculate the limit value of the second using polar co-ordinates, by taking  $x=r\cos\theta$ and $y=r\sin\theta$ and $r\rightarrow 0.$
Therefore $g$ is  differentiable at the origin.
