Does the surface $f(x,y)=\sqrt{|xy|} $ have a tangent plane at origin? $f(x,y)=\sqrt{|xy|} $ is a level set of $g(x,y,z)=f(x,y)-z$ and now I claim that partial derivatives of $f(x,y)$ exist at zero. 
$\lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{0-0}{h}=0$ and similarly, $\lim_{h\to 0}\frac{f(0,0+h)-f(0,0)}{h}=\lim_{h\to 0}\frac{0-0}{h}=0$ and hence $f_x(0,0)=0=f_y(0,0)$ 
$\nabla g(0,0,0)=(g_x(0,0,0),g_y(0,0,0),-1)=(0,0,-1)$ 
Hence, equation of tangent plane at origin is: $0(x-0)+0(y-0)+(-1)(z-0)=0\implies z=0$ 
I believe that the above is correct. But I am confused because as per answer sheet, the surface $f(x,y)=\sqrt|xy|$ doesn't have tangent plane at origin. Please help. Thanks.
Edit: A typo ($f(x, y) =|xy|$ was taken instead of $f(x, y) =\sqrt{|xy|} $) inadvertently crept in in the earlier version of this question and the same has been edited.
 A: As I see for $f(x,y)=\left|xy \right|$ your calculation of partial derivatives is correct, and function is  differentiable at $(0,0)  $. And more, partial derivatives are continuous:
$\left(\left| xy\right|\right)_{x}^{'}=\left| y\right| \cdot sgn(x)$.
Also by direct calculation :
$$\frac{\left|xy \right|}{\sqrt{x^2+y^2}} \leqslant \left|x \right|$$
So it is  differentiable and have tangent in $(0,0)$.
Note:
If we take function $g(x,y)=\sqrt{\left|xy \right|}$, then it's possible to calculate,that it's partial derivatives exists in $(0,0)$, but are not bounded: $g_{x}^{'} = \frac{1}{2} \sqrt{\left| \frac{y}{x} \right|}$. For differentiablity if we take $(\frac{1}{n},\frac{1}{n})$, then it will be
$$\frac{\sqrt{\left|xy \right|}}{\sqrt{x^2+y^2}} =\frac{1}{\sqrt{2}}$$
So, for $g$ we have not tangent - Your book is correct: Apostol is Apostle.
A: Allawonder's answer is correct, in a more detailed manner, you might see that if you view the graph the surface is $0$ along the main axis, which is the axis you
consider when doing the partial derivative. In other words, the partial derivative
is a directional derivative along the standard basis vectors of your space, $(0,1)$ and $(1,0)$.

However, a tangent plane, should it exist, can be defined for any choice of basis vectors by taking directional derivatives. If you were to take these directional derivatives for the vectors $(1,1)$ and $(1,-1)$ you would immediately run into a problem when trying to apply the formula as you did in your
example. Try calculating
$$\lim_{h \to 0} \frac{f(0 + h, 0 + h) - f(0, 0)}{h},$$
and you should see why you can not have a tangent plane in this point.
