Here is a tedious answer using the implicit function theorem applied to $\phi(X,Y) = X-Y^2 = 0$, with $X,Y \in \mathbb{C}^{n \times n}$.
It is straightforward to see that $\frac{\partial \phi(X,Y)}{\partial X}\Gamma = \Gamma$, and $\frac{\partial \phi(X,Y)}{\partial Y}\Delta = -(\Delta Y + Y \Delta)$. Since both are smooth, it follows that $\phi$ is smooth as well.
We need to show that $L=\frac{\partial \phi(X,Y)}{\partial Y}$ is invertible, at least in a neighborhood of a Hermitian $Y>0$.
If $Y>0$ and is Hermitian, it has a full set of eigenvectors $v_n$ corresponding to each (real) eigenvalue $\lambda_n$. It is easy to check that $L(v_n v_m^*) = -(\lambda_n+\lambda_m)v_n v_m^*$, and that $\{v_n v_m^*\}_{m,n}$ is a basis. Since $\lambda_n+\lambda_m>0$, for all $m,n$, $L$ is invertible. Furthermore, note that if $B$ is Hermitian, then the solution $\Delta$ to $L(\Delta) = B$ will also be Hermitian.
Consequently, if we have $\phi(\hat{X},\hat{Y}) = 0$, with Hermitian $\hat{X},\hat{Y} >0$, then there exists a unique $\eta$ defined in a neighborhood of $\hat{X}$, such that $\eta(\hat{X}) = \hat{Y}$, and $\phi(X,\eta(X)) = 0$, for $X$ in this neighborhood. Furthermore, $\eta$ is differentiable on this neighborhood.
However, we are not finished. It remains to show that if $X>0$ is Hermitian (and sufficiently close to $\hat{X}$ so that $\eta(X)$ remains positive definite), then $\eta(X)$ is Hermitian. Let $\Xi(t) = \hat{X}+t(X-\hat{X})$, and $H(t) = \eta(\Xi(t))$. Then $\dot{H}(t) = \frac{\partial \eta(\Xi(t))}{\partial X} (X-\hat{X}) = \frac{\partial \phi(\Xi(t),H(t))^{-1}}{\partial Y} (X-\hat{X})$, with $H(0) = \hat{Y}$, of course. If we let $f(H,t) = \frac{\partial \phi(\Xi(t),H)^{-1}}{\partial Y} (X-\hat{X})$, then this can be written as $\dot{H} = f(H,t)$, $H(0) = \hat{Y}$. Note that if $H$ is Hermitian, then $f(H,t)$ is also Hermitian.
Now let $\Pi$ be the projection onto the subspace of Hermitian matrices, and consider the equation $\dot{J} = \Pi f(J,t)$, $J(0) = \hat{Y}$. Since $J(t) = J(0) + \int_0^t \Pi f(J(\tau),\tau) d \tau = \Pi(J(0) + \int_0^t f(J(\tau),\tau) d \tau)$, we see that $J(t)$ is Hermitian. It follows from this that $\dot{J} = f(J,t)$, and by uniqueness of solution we have that $H(t) = J(t)$, hence $H(t)$ is Hermitian, and so $\eta(X) = H(1)$ is Hermitian.