Checking differentiability of a function Let
$$f(x,y,z)=\left\{\begin{array}{c c}\dfrac{\ln(1+x^2y^4z^6)}{(x^2+y^2+z^2)^\alpha} & \textrm{if } (x,y,z)\neq (0,0,0) \\ 0 &\textrm{if } (x,y,z)=(0,0,0)\end{array}\right. $$
I want to know for which values for $\alpha$ this function is differentiable.
Since, for every $(x,y,z)\in\Bbb{R}^3,\; 1+x^2y^4z^6>0,$ and $x^2+y^2+z^2>0$ if $(x,y,z)\neq(0,0,0)$.  So, $f$ is differentiable for every $(x,y,z)\neq(0,0,0).$
Then I just need to check for $(0,0,0).$
If $\alpha\leq 0,$ the denominator will not vanish, so $f$ will be differentiable... but I don't know what happens if $\alpha >0$. Any tips?
 A: How about this. Fix $u_1,u_2,u_3\in \mathbb{R}$ such that $u_1^2+u_2^2+u_3^2=1.$ Define the function $l(t)$ by $$l(t)=f(tu_1,tu_2,tu_3)$$It turns out $l'(0)$ is the directional derivative of $f$ along $u=\big<u_1,u_2,u_3\big>$ at the origin, namely $D_{u}f(0,0,0)$. You need to find all $\alpha\in\mathbb{R}$ that makes $\lim_{t \rightarrow 0}\frac{l(t)}{t}$ exist for any choice of $u_1,u_2,$ and $u_3$. Notice if either $u_1,u_2,$ or $u_3$ equals $0$ then $l\equiv0$ and so $l'(0)=0$ obviously. On the other hand, if neither $u_1,u_2$ nor $u_3$ equals $0$, then we have for $t\neq0$ that$$\frac{l(t)}{t}=\frac{\ln(1+u_1^2u_2^4u_3^6t^{12})}{u_1^2u_2^4u_3^{6}t^{12}}\cdot \frac{u_1^2u_2^4u_3^6t^{12}}{t^{2\alpha+1}}$$
The limit as $t \rightarrow 0$ exists in the above expression whenever $\alpha \leq 11/2$.
PS: In case you're wondering, $D_uf(x_0,y_0,z_0)=l'(0)$ where $$l(t)=f(x_0+u_1t,y_0+u_2t,z_0+u_3t)$$
A: If we take $y=x=z$, then we obtain $\frac{\ln (1+x^{12})}{(3x^2)^\alpha}$, so for $\alpha \geqslant 6$ we have discontinuity in $(0,0,0)$.
