Following is my cost that I need to minimize wrt $\mathbf{y}$
\begin{equation} J = (\mathbf{y^T\mathbf{z_1}})^2+(\mathbf{y^T\mathbf{z_2}})^2-\lambda(\mathbf{y}^T\mathbf{e}-1) \end{equation}
$\lambda$ is a scalar varaible.
My work so far
\begin{align} J &= (\mathbf{y^T\mathbf{z_1}}\mathbf{z_{1}^{T}\mathbf{y}})+(\mathbf{y^T\mathbf{z_{2}}}\mathbf{z_{2}^{T}\mathbf{y}})-\lambda(\mathbf{y}^T\mathbf{e}-1)\\ &= (\mathbf{y^T\mathbf{A_1}\mathbf{y}})+(\mathbf{y^T\mathbf{A_{2}}\mathbf{y}})-\lambda(\mathbf{y}^T\mathbf{e}-1) \end{align}
\begin{align} \frac{\partial J}{\partial \mathbf{y}} &= \left( (\mathbf{A_1}+\mathbf{A_1^{T}})\mathbf{y} + (\mathbf{A_2}+\mathbf{A_2^{T}})\mathbf{y} -\lambda\mathbf{e}\right)\\ &= \left( (\mathbf{A_1}+\mathbf{A_1^{T}} + \mathbf{A_2}+\mathbf{A_2^{T}})\mathbf{y} -\lambda\mathbf{e}\right)=0 \end{align} this leaves \begin{align} \mathbf{y} &= (\mathbf{A_1}+\mathbf{A_1^{T}}+\mathbf{A_2}+\mathbf{A_2^{T}})^{-1} (\lambda\mathbf{e}) \end{align}
My problem is with finding $\lambda$
Notice that $A_1$ and $A_2$ are symmetric, since $(zz^T)^T=(zz^T)$.
So going back to the derivative wrt $y$, you can write $$ \eqalign{ \lambda e &= 2\,(A_1+A_2)y \cr }$$ Next, find the gradient with respect to $\lambda$ and set it to zero $$ \eqalign{ \frac{\partial J}{\partial\lambda} &= y^Te-1 &= 0 \cr 1 &= y^Te \cr }$$ and substitute this into the first equation (multiplied by $y^T$) $$ \eqalign{ \lambda y^Te &= 2\,y^T(A_1+A_2)y \cr \lambda &= 2\,y^T(A_1+A_2)y \cr }$$ Finally, you can substitute this expression for $\lambda$ into your result to obtain the optimal $y$.