I think there was an issue with the question I asked earlier.
I want to prove that the quaternion is a hyperkahler manifold. I know that there is a natural metric $\rho$ on that given by $\rho(a,a)=a·a^{*}$ which is just the product in $\mathbb{H}$ which gives a manifold structure to $\mathbb{H}$ which is diffeomorphic to $\mathbb{R}^{4}$.
So how could we constructer the hyperkahler structure?
Of course, for any $q \in \mathbb{H}$, $T_q\mathbb{H} \simeq \mathbb{H}$, and consider as complex structures $I : q \mapsto iq$ and $J: q \mapsto jq$. Of course these anticommute and $I^2 = J^2 = -id$. Moreover, they are obviously isometries with respect to $\rho$ and both are orientation preserving. Indeed, if $\{1, i, j, k\}$ is a basis for $\mathbb{R}^4 \simeq \mathbb{H}$, the matrix expressions of these maps are $$ I = \begin{bmatrix} 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1\\ 0 & 0 & 1 & 0 \end{bmatrix} \quad J = \begin{bmatrix} 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 \end{bmatrix} $$ which you can easily verify to have determinant $1$.
This is the natural hypercomplex structure on $\mathbb{R}^4$ induced by an isomorphism with $\mathbb{H}$, to prove this structure is hyperkähler one needs to show that both $I,J$ are parallel with respect to the Levi-Civita connection. But on $\mathbb{R}^4$ this is trivial as, with respect to the Euclidean metric, its Levi-Civita connection is flat. Indeed, if $q : \mathbb{R}^4 \to \mathbb{R}^4$ is a vector field $$ (\nabla I)q = \nabla(Iq) - I \nabla q = d(Iq) - I d q = d I q + I dq -I d q = 0 $$ as $I$ is constant. As the same can be done for $J$, one concludes $\nabla I = \nabla J = 0$.