There is no such function.
Every function that has partial derivatives is obviously continuous in $x$ for fixed $y$ and vice versa, so it must be of first Baire class, hence continuous almost everywhere in the category sense.
To prove this, we may "mollify" $f$ in $x$ for every fixed $y$ - consider, e.g., $f_\varepsilon := f \ast_x \varphi_\varepsilon$, where $\varphi_\varepsilon$ is a smooth compactly supported function of one variable, and $\ast_x$ is convolution with respect to $x$. Since $\varphi_\varepsilon$ is smooth, $f_\varepsilon(\cdot,y)$ are equicontinuous on compact sets; since they also inherit from $f$ its separate continuity, they become actually continuous in both variables. On the other hand, if $\varphi_\varepsilon$ approaches $\delta_0$, $f_\varepsilon$ converge to $f$ pointwise, so $f$ is indeed of first Baire class.
After having a look at Lemma 13.11 in your textbook, I think you are confusing the application of the mean value theorem in one dimension with the one in higher dimensions.
I've never seen that lemma before, so I will try to explain the proof here, if only to have a reference without having to look for the exact page in the book again.
Examine the second-order differences
$$f(x_0 + r, y_0 + r) - f(x_0 +r, y_0) - f(x_0, y_0 +r) + f(x_0, y_0) \quad (1)$$
for some small positive $r$. You could have different displacements $r_1$ and $r_2$ for the two coordinates, positive or negative, but with small absolute value each, but choosing $r_1 = r_2$ is notationally more convenient (but try the proof with different displacements, it makes for an easy exercise). In any case, you want to get second-order derivatives from that. The mistake that I made when looking at this first was to apply the mean value theorem for $f(x_0+r,y_0+r) - f(x_0+r,y_0)$ in the second variable and likewise for the other difference. But this will lead you to an impasse (think about why). The trick is to consider the functions $$x \mapsto f(x,y_0+r) - f(x,y_0) \quad \text{and}\quad y \mapsto f(x_0+r,y) - f(x_0,y)$$ and apply the mean value theorem to them. This works because $f$ is supposed to have partial derivatives up to order 2 in all directions, so it is continuously differentiable (partially in both variables, if you like that equivalent statement better), which is more than we need to apply the MVT. We get that the differences (1) is equal to both$$r\left(\frac{\partial f}{\partial x}(x_1,y_0 +r) - \frac{\partial f}{\partial x}(x_1,y_0)\right) \quad \text{and} \quad r\left(\frac{\partial f}{\partial y}(x_0+r,y_2) - \frac{\partial f}{\partial y}(x_0,y_2)\right), \quad (2)$$
for some $x_1 \in [x_0,x_0+r]$ and $y_2 \in [y_0,y_0+r]$. But the function $y \mapsto \frac{\partial f}{\partial x}(x_1,y) - \frac{\partial f}{\partial x}(x_1,y)$ is continuous on $[y_0,y_0+r]$ and differentiable on the interior, which is all we need to apply the mean value theorem again. In particular, the derivative need not be continuous! A similar function does the job for the second difference in (2). So (1) is equal to
$$r^2 \frac{\partial^2 f}{\partial y \partial x}(x_1,y_1) \quad \text{and} \quad r^2 \frac{\partial^2 f}{\partial x \partial y}(x_2,y_2)$$
for some $y_1,x_2$ in the right intervals. Dividing by $r^2$ we get the result.
Added: See also “The Mixed Partial Derivatives and the Double Derivative” by Donald H. Trahan, The American Mathematical Monthly , Vol. 76, No. 1 (Jan., 1969), pp. 76-77 (JSTOR link)
Best Answer
HINT: You should think about two different paths. One path is the line segment from $(a,b)$ to $(a+h,b+k)$. The other path is the two-segment path, going first in the $y$-direction from $(a,b)$ to $(a,b+k)$ and then in the $x$-direction from $(a,b+k)$ to $(a+h,b+k)$.