Given
$$ \alpha\Delta^2y+y=\alpha^{1/4}y_d-\alpha^{5/4}\Delta(f+u_d), \Omega $$
$$ y=0, \alpha^{1/2}\Delta y+\alpha^{3/4}(f+u_d)=0 $$
I am trying to verify the regularity result as
$$ \alpha^{1/2}|y|_{H^2}+\|y\|_{L^2}\leq C\alpha^{1/4}(\|y_d\|_{L^2}+\alpha^{1/2}\|f+u_d\|_{L^2}) $$
The hint is to use the variational form but I am not sure how to proceed with
$$ (\Delta^2y,v)=(\Delta y\Delta y,v)=\int\frac{\partial y}{\partial n}\Delta y v -? $$
Here is an incomplete answer providing the weak form. The regularity result is still giving me trouble. Let $y,v\in H^2_0(\Omega)$ and $(\cdot,\cdot) = (\cdot,\cdot)_{L^2(\Omega)}$. Then Green's identities yield
$$ \begin{aligned} (\Delta^2y,v) = (\Delta y, \Delta v) + \int_{\partial\Omega} v(\nabla\Delta u\cdot\hat{n}) - \Delta y(\nabla v\cdot \hat{n})ds. \end{aligned} $$
Applying this to the original PDE and applying similar Green's identities to the other terms, we obtain
$$ \alpha\left((\Delta y, \Delta v) + \int_{\partial\Omega} v(\nabla\Delta u\cdot\hat{n}) - \Delta y(\nabla v\cdot \hat{n})ds\right) + (y,v) = \alpha^{1/4}(y_d,v) - \alpha^{5/4}\left((f+u_d,\Delta v) + \int_{\partial\Omega}v(\nabla(f+u_d)\cdot\hat{n}) - (f+u_d)(\nabla v\cdot\hat{n})ds\right). $$ Collecting like terms, we obtain
$$ \alpha(\Delta y,\Delta v) + (y,v) = \alpha^{1/4}(y_d,v)+\alpha^{5/4}( f+u_d,\Delta v) \\ +\int_{\partial\Omega}\left(\alpha\Delta y +\alpha^{5/4}(f+u_d)\right)(\nabla v\cdot\hat{n}) + v\left(\alpha\nabla\Delta u\cdot\hat{n} - \alpha^{5/4}\nabla(f+u_d)\cdot\hat{n}\right). $$
Because of the boundary condition $\alpha^{1/2}\Delta y + \alpha^{3/4}(f+u_d) = 0$, the first term in the boundary integral is zero. The second term is zero because $v\in H^2_0(\Omega)$. Therefore, your weak form reduces to $$ \alpha(\Delta y,\Delta v) + (y,v) = \alpha^{1/4}(y_d,v)+\alpha^{5/4}( f+u_d,\Delta v) \quad \forall v\in H^2_0(\Omega). $$
It's a bit late for me so I'm missing how to prove the bound right now but letting $v=y$ then using CS seems like it yields something like what you want.