If $h$ is a diffemorphism, then $dh$ is everywhere nonsingular. Because
$$
d(f \circ h) = df \circ dh
$$
(depending on notation, etc. --- we're talking about the chain rule here), we have $d(f\circ h)(P)(v)$ is zero exactly when $df(Q)(w)$ is zero (where $Q = h(P)$ for some nonzero $w$, because $dh(P)(v)$ is never zero for any nonzero $v$, because $h$ is a diffeomorphism.
The one weird thing about this question is that "isotopic to the identity" is completely irrelevant; is there a part "b" or "c" that might use that assumption?
Here is a somewhat sketchy answer, based on the creed that Morse functions and (smooth) handles should be thought about interchangeably (and that Heegaard splittings are a form of handle decomposition).
Suppose $M$ is a closed, connected, orientable 3-manifold and $f:M\rightarrow [0,3]$ is a self-indexing Morse function. For $k=0,1,2,3$, let $c_k$ be the number of critical points of $f$ having index $k$. We can see that $c_1-c_0=c_2-c_3$, because $$0=\chi(M)=c_0-c_1+c_2-c_3.$$
Since $f^{-1}[0,1.5]$ is formed from $c_0$ zero-handles and $c_1$ one-handles, we can see that it is a handlebody of genus $c_1-c_0+1$ (this uses the assumption that $M$ is connected and orientable). Considering the upside-down Morse function $-f$, we can similarly observe that $f^{-1}[1.5,3]$ is a handlebody of genus $c_2-c_3+1$. So the genus of this particular Heegaard splitting is $c_1-c_0+1=c_2-c_3+1.$ If $c_0=c_3=1$, then the genus of this particular Heegaard splitting is $c_1=c_2.$
Suppose that our Morse function is minimal, i.e. it has $\mathfrak m(M)$ critical points. If we have $c_0>1$, then there are multiple zero-handles, which need to be glued together by 1-handles in order for $M$ to be connected. Hence, there is an index-one critical point $q$ whose two descending flow lines limit to two different critical points $p$ and $p'$ of index zero. Since there is exactly one flow line from $q$ to $p$, we can cancel these critical points, which contradicts the minimality of $f$. Thus, we must have had $c_0=1$ to begin with. We can see that $c_3=1$ by applying the same argument to $-f$. This proves that $$\mathfrak h(M)\leq c_1=\frac{2c_1+2}{2}-1=\frac{c_0+c_1+c_2+c_3}{2}-1=\frac{\mathfrak m (M)}{2}-1.$$ To see the reverse inequality, we need to prove that a genus $g$ Heegaard splitting gives rise to a Morse function $f:M\rightarrow [0,3]$ with $c_0=c_3=1$ and $c_1=c_2=g$. Letting $H_g$ denote a handlebody of genus $g$, we can define a self-indexing Morse function $h:H_g\rightarrow[0,1.5]$ having one index-zero critical point and $g$ index-one critical points, and satisfying $h^{-1}(1.5)=\partial H_g$ (because $H_g$ can be formed from a single zero-handle and $g$ one-handles). Then $3-h:H_g\rightarrow[1.5,3]$ is also a self-indexing Morse function having one index-three critical point and $g$ index-two critical points, and satisfying $(3-h)^{-1}(1.5)=\partial H_g$. Being careful to preserve smoothness, we can glue these two functions along the splitting surface of the Heegaard splitting to define the desired Morse function $f:M\rightarrow [0,3].$
Best Answer
Initially I was looking for something that used a little less machinery, but the problem is pretty easy once we assume Whitney embedding.
Let $f: M \to \mathbb R^m$ be an embedding, which we can find for $m$ sufficiently large, so now we just think of $M \subseteq \mathbb R^m$. We can translate $M$ so that it does not hit the origin, in which case the norm map $g: x \mapsto |x|$ is smooth on $M$ and proper. By theorem in Guillemin and Pollack, the map \begin{equation*} g_a = g + a \cdot x \end{equation*} is Morse for almost ever $a \in \mathbb R^m$. Choose some $a$ with sufficiently small norm, say $|a|<1/2$, such that $g_a$ is Morse, then $|g_a (x)| \geq |x|/2$, so $g_a$ is also proper.
I guess the question still stands, is there a way of doing this without Whitney embedding?