An intrinsically curved surface with no extrinsic curvature An intrinsic curvature is the property of a manifold itself. It shows it's deviation from euclidean geometry. The extrinsic curvature, on the other hand, depends on how the surface is embedded in a higher dimensional manifold. Now as a beginner student wherever I've seen examples of difference between these two I'm always led to the example of a cylinder. It has no intrinsic curvature but looking at it from R3 it does have an extrinsic curvature.
My question is : Is the opposite true? Can I have a surface with no extrinsic curvature(due to my "clever" embedding) but is intrinsically curved? If yes, what will be a simple example?
 A: Let's first clarify the terminology and setting. Let's assume that we have an embedded submanifold $S$ (e.g. a surface) in a Riemannian manifold $M$ (e.g. Euclidean 3-space).
I assume that by intrinsic curvature of $S$, you mean the Gaussian curvature of a surface. For higher dimensional manifolds, this generalizes to the sectional curvature, but this is a little more complicated: it assigns a number to each 2-dimensional subspace of the tangent space, namely the Gaussian curvature of the submanifold (surface) tangent to that plane.
"Extrinsic curvature" could mean several things, but for a hypersurface (e.g. an embedded surface $S$ in Euclidean 3-space), we could summarize by saying that an extrinsic curvature is a quantity defined by the second fundamental form, or equivalently its associated shape operator $B$. If you don't know what these things are, it's ok, you can keep reading. Think of the shape operator as a symmetric matrix depending on $p \in S$. The main "extrinsic curvatures" that are worth considering are:


*

*The eigenvalues of $B$, called principal curvatures. Note that they are equal to the curvature of curves lying in $S$, seen as curves in $M$.

*The trace of $B$ (maybe divided by the dimension), called the mean curvature, equal to the sum (or average) of the principal curvatures.


For a surface in a 3-dimensional manifold, the Gauss equation says that:
$$ \det B = K_S - K_M$$
where $K_S$ is the Gauss curvature of $S$ and $K_M$ is the sectional curvature in $M$ of the plane tangent to $S$. This equation is probably one of things you're looking for answering your question: it tells you the relation between the second fundamental form (defining the "extrinsic curvatures"), the intrinsic curvature of $S$ and the intrinsic curvature of $M$.
Now let's answer your question more precisely. As you can tell from Gauss equation, if all the extrinsic curvatures are zero, i.e. $B$ vanishes (FYI, in this case one says that $S$ is a totally geodesic submanifold), then the intrinsic curvature of $S$ is equal to the intrinsic curvature of $M$. In particular, if $M$ has zero curvature (i.e. Euclidean 3-space), then a submanifold for which all the extrinsic curvatures are zero also has zero intrinsic curvature.
That being said, if by "the extrinsic curvature" we just signify the mean curvature $\mathrm{tr}(B)$, we are just looking for surfaces with zero mean curvature, these things are called minimal surfaces. Now the question is: are there minimal surfaces that are not totally geodesic (basically, minimal surfaces that are not planes in Euclidean 3-space)? You can guess that the answer is probably "yes, there are plenty", because that's pretty much asking if there are some symmetric matrices $B$ whose trace is zero, but are not the zero matrix. In fact, the "fundamental theorem of surface theory" basically guarantees that there are many examples. You'll see examples by looking for images of "minimal surface" on the web.
I tried to give a complete and detailed answer, I hope it was useful, but if you're looking for the short answer: yes, any minimal surface in Euclidean 3-space that is not a plane has zero mean curvature but nonzero Gaussian curvature. For example, the catenoid:

