The function might be the world’s most popular example demonstrating that the existence of partial derivatives does not imply differentiability.

But in my opinion, it is somewhat extreme and potentially confusing, with discontinuity added to the mix. I prefer
pictured below.

This one is continuous. In fact, it is Lipschitz continuous because the first-order partials and
are bounded. The restriction of
to the line
is
, which is a familiar single-variable example of a nondifferentiable function.
To unify the analysis of such examples, let . Then
With , where
, we get
By symmetry, is bounded as well.
My favorite example from this family is more subtle, with a deceptively smooth graph:

The formula is
Since decays almost quadratically near the origin, it is differentiable at
. Indeed, the first order derivatives
and
are continuous, as one may observe using
above.
And the second-order partials and
are also continuous, if just barely. Indeed,
Since the growth of is sub-logarithmic, it follows that
and
. Hence,
So, as
. Even though the graph of
looks quite similar to the first example in this post, this one is continuous. Can’t trust these plots.

By symmetry, is continuous as well.
But the mixed partial does not exist at
, and tends to
as
. The first claim is obvious once we notice that
and
blows up at
. The second one follows from
where while the other two terms tend to zero, as in the estimate for
. Here is the graph of
.

This example is significant for the theory of partial differential equations, because it shows that a solution of the Poisson equation with continuous
may fail to be in
(twice differentiable, with continuous derivatives). The expected gain of two derivatives does not materialize here.
The situation is rectified by upgrading the continuity condition to Hölder continuity. Then indeed gains two derivatives: if
for some
, then
. In particular, the Hölder continuity of
and
implies the Hölder continuity of
.