Differentiability of a scalar field 
Is $f(x,y,z)=x\sqrt{y^2+z^2}$ differentiable at the point $(0,0,0)$? Prove your assertion.

I'm pretty sure this function is actually differentiable at $0$. The problem is I'm unsure how exactly I should go about proving this. My understanding is that if the partial derivatives all exist at zero and are also continuous about zero, this would be enough to prove differentiability. How would I show these partial derivatives are continuous?
Note that the partial derivatives are all equal to zero at the origin.
With partial derivatives defined as:
(1) $\frac{\partial f}{\partial x}=\sqrt{y^2+z^2}$
(2) $\frac{\partial f}{\partial y}=\frac{xy}{\sqrt{y^2+z^2}}$
(3) $\frac{\partial f}{\partial z}=\frac{xz}{\sqrt{y^2+z^2}}$
 A: Do you know if $f$ is $\textit{supposed}$ to be differentiable at the origin? (I guess this is implied by your inclusion of the phrase 'prove this assertion', but could you confirm?)  If you don't, then it's possible that $f$ isn't in fact differentiable at the origin, and simply showing that some of the partial derivatives are not continuous at the origin is not sufficient to prove non-differentiability.  You would have to prove that the tangent plane to the graph of $f$ is not a good linear approximation to the graph of $f$ at $(0,0,0)$.
A: The derivative of a scalar field is the vector function defined by the gradient of that field. In order for $f$ to be differentiable, then the gradient would have to exist at the point in question. Find the gradient and then evaluate at $(0,0,0)$ to see if it exists.
I believe that $f_y$ and $f_z$ do not exist at $(0,0,0)$. Because these values do not exist, those functions are not continuous at the origin and therefore $f$ is not differentiable at this point. 
