At my undergraduate institution (Facultad de Ciencias, UNAM, Mexico), for a while in the mid-to-late 70s, several professors in the Calculus sequence (four courses: Differential single-variable (Calc I), Integral single-variable (Calc II), Differential multi-variable (Calc III), Integral multi-variable (Calc IV)) decided to use Hasse's analysis textbook instead of a calculus textbook. It was more of a "baby analysis" than a calculus course.
Now, this was done only in courses that were being taught to Math, Actuarial Sciences, and Physics majors (and a Math major takes nothing but math courses, for instance).
It did not go well. Students didn't learn analysis very well, and they certainly did not learn the calculus skills they needed very well. The Physics department, in particular, went up in arms because the Physics majors were coming out of these courses unable to actually compute integrals and derivatives, or use them to solve specific physics problems. Same problem with the actuarial scientists. The math majors fared a little better, but mainly because the same people who were doing this were the people who were also teaching the analysis courses in the junior and senior years; but those that went on to take analysis from other people didn't do so well. In addition, the failure rate for these courses was extremely high. (Failure rate in the Calculus sequence has always been way too high there, but it got much worse).
Most professors switched back to calculus books and to not do baby analysis. By the mid-80s, almost nobody was using Hasse's book or teaching "mini-analysis."
If a student has had a good enough calculus course in High School, then it is likely that a baby analysis course might indeed be beneficial, building on the bases that calculus can help set. This could very well be the case in the EU; it's not the case in the US. (In Mexico, nominally, students in the Math/Physics/Engineering track were taking a year of Calculus as seniors in High School, but obviously not good enough).
A function $f:\mathbb{R}^2 \to \mathbb{R}$ is differentiable at a point $p$ if there is a linear map $L:\mathbb{R}^2 \to \mathbb{R}$ such that $$\lim_{h \to 0}\frac{\left|f(p+h)-f(p)-L(h)\right|}{|h|} = 0$$ If the function is differentiable, then $L$ is the function $L(x,y) = \dfrac{\partial f }{\partial x}\big|_p x +\dfrac{\partial f}{\partial y}\big|_p y$.
If all you know is that the partial derivatives exist, you do not even know that the function is continuous, let alone is well approximated by a tangent plane. For instance $f(x,y) = 0$ if $x=0$ or $y=0$ and $f(x,y) = 1$ otherwise. The partial derivatives of $f$ at $(0,0)$ are all $0$, but the tangent plane is a really crappy approximation to $f$ off of the coordinate axes.
The example you were looking at is a little harder to visualize, but think about what happens to $f$ on lines other than horizontal and vertical ones.
In other words, morally, for a function to be differentiable at $(a,b)$ we need that $f(a+\Delta x,b+\Delta y) \approx f(a,b) + \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y$. This talks about approximating $f$ in all directions around $(a,b)$, whereas existence of the partial derivatives only means you have good approximations in two directions (the coordinate directions).
Best Answer
Try $$f(x,y)=e^{1/(x^2+y^2)} x y, \qquad f(0,0)=0$$ Away from $(0,0)$ it is smooth, but at $(0,0)$ only the partial derivatives in directions $(1,0)$ and $(0,1)$ exist, not in the direction $(1,1)$. The $\frac{\partial^2 f}{\partial x \partial y}$ second derivative doesn't exist neither.
Another example would be
$$g(x,y)=xy\sin(\frac1{x^2+y^2})$$ It is smooth away from $(0,0)$.
$g(x,y)=O(x^2+y^2)$ so the derivatives in every directions exist and vanish at $(0,0)$, but $(\partial_x g)(0,y)=y\sin(1/y^2)$ which is not differentiable at $y=0$ so that $(\partial_y (\partial_x g))(0,0)$ doesn't exist.