As CamSar notes, if you assume that $Y$ is smooth, the result is true. In fact, if you even just assume that $Y$ is normal and that $df$ is an isomorphism at the generic points, then Zariski's Main Theorem implies that $f$ is an isomorphism if $X$ is reduced and $f$ is bijective. As Nakajima explains on the top of p. 12, his space $\mathfrak{M}(\mathbf{v},\mathbf{w})$ is smooth. Moreover, the group action is by the linearly reductive group $\mathbb{G}_m$. Thus the fixed point set $Y$ is smooth by "Iversen's theorem". So that explains the result in Nakajima's paper.

Of course there are counterexamples when $Y$ is singular. If you allow $X$ and $Y$ to be nonreduced, it is trivial to construct examples, as explained in the answers here. However, your wording suggests that you are interested in the case when $X$ and $Y$ are integral varieties. None of the previous answers seems to address that case. So, just for the record, following is an example where $X$ and $Y$ are integral varieties, $f$ is bijective and isomorphic on Zariski tangent spaces, yet $f$ is not an isomorphism. Note, the following morphism $f$ is **not** finite -- if it were, then the theorems mentioned in your comment would imply that $f$ is an isomorphism.

Beginning with $\mathbb{A}^2$ with coordinates $(x,y)$, let $Y$ be the plane curve $$ Y = \{(x,y)\in \mathbb{A}^2 : y^4+y^2x-x^3 = 0 \}.$$ This is an irreducible curve with a unique singular point at $(x,y)=(0,0)$.
Next, beginning with $\mathbb{A}^3$ with coordinates $(u,v,w)$, let $X$ be the curve $$X = \{(u,v,w)\in \mathbb{A}^3 :(v+1)w-1 = v^3+v^2-u^2=0\}. $$

Projection from $X$ to the $(u,v)$-plane is a locally closed immersion with image the complement of $(0,-1)$ in the plane curve with equation $v^3+v^2-u^2$. In particular, $X$ is an irreducible curve with a unique singular point at $(0,0,1)$. Now consider the morphism,
$$ F :\mathbb{A}^3 \to \mathbb{A}^2, \ \ f(u,v,w) = (u,v^2+v).$$

I claim that $F(X)$ equals $Y$, and the restriction morphism,
$$ f :X \to Y,$$
is a bijection that induces isomorphisms on all Zariski tangent spaces. The simplest way to prove this is to first normalize $X$ and $Y$. In the function field of $X$, observe that the following monic equation holds,
$$ t^2 - (v+1) = 0, \ \ t = u/v.$$
Thus, $k[t]$ is a subring of the integral closure of the fraction field. But, of course, $v = t^2 - 1$ and $u = t(t^2-1)$ are already in $k[t]$. Finally, $w = 1/t^2$ is in the integral closure. Thus, the integral closure of $k[X]$ equals the integral closure of $k[t][1/t^2]$, which is already the integrally closed ring $k[t][1/t]$. So the normalization of $X$ is just $$\nu: \mathbb{G}_m \to X, \ \ \nu(t) = (t(t^2-1),t^2-1,1/t^2).$$

In particular, the composition with $F$ is,
$$ F\circ \nu: \mathbb{G}_m \to \mathbb{A}^2, \ \ F(\mu(t)) = (t(t^2-1),t^2(t^2-1)).$$ By the same argument as above, the normalization of $Y$ is,
$$ \mu: \mathbb{A}^1 \to Y, \ \mu(t) = (t(t^2-1),t^2(t^2-1)).$$
Thus, $F\circ \nu$ is just the restriction of the normalization $\mu$ to the open $\mathbb{G}_m\subset \mathbb{A}^1$.

Since $F\circ\nu$ factors through the normalization of $Y$, in particular $F(X)$ is contained in $Y$. Denote by $f$ the restriction. Moreover, since $\nu$ is a bijection except over $(0,0,1)$, also $f$ is injective at that point. Similarly, since the domain of $F\circ \nu$ contains every point except $t=0$, the image of $f$ contains every point, except possibly $\mu(0)=(0,0)$. But by direct computation, the inverse image under $f$ of $(0,0)$ is precisely $(0,0,1)$. Therefore $f$ is bijective.

Since $f$ matches up normalizations, $f$ is an isomorphism at every point where the normalization morphisms $\nu$ and $\mu$ are isomorphisms, i.e., everywhere except $(0,0)$ in $Y$ and $(0,0,1)$ in $X$. In particular, except possibly at those points, $df$ is an isomorphism. Finally, by direct computation, $df$ is an isomorphism at the point $(0,0,1)$: both Zariski tangent spaces are two dimensional, and $df$ is an isomorphism between them (just linear projection, really).

Finally, $f$ is **not** an isomorphism. If $f$ were an isomorphism, it would induce an isomorphism of normalizations. But the induced morphism of normalizations is the non-surjective inclusion, $\mathbb{G}_m\subset \mathbb{A}^1$.