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 $g(x,y)$.
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.