The Schwarzschild solution could simply be expressed as
$$ds^2=-(1-2GM/r)dt^2+(1-2GM/r)^{-1}dr^2+r^2d\phi^2 \; .$$
Is it possible that we could obtained a new metric into the form as
$$ds^2=-(1-2GM/r)^{-1}dt^2+(1-2GM/r)dr^2+r^2d\phi^2 \; ?$$
If possible what are the steps and procedures that should be done to derive it in formal way.
The Schwarzschild metric can be derived requiring spherical symmetry and vacuum solution ($R_{\mu\nu} = 0$) for the metric: $$ds^2 = -Adt^2 + Bdr^2 + Cr^2d\theta^2 + Dr^2sin^2(\theta)d\phi^2$$ where A, B, C and D are functions of $t$,$r$, $\theta$ and $\phi$
You can calculate the Christoffel symbols and the components of the Ricci tensor for the alternative metric. After that, verify if it can be a vacuum solution. If not, what is the energy-momentum tensor associated.
Your other metric is not a solution to Einstein field equation. The metric coefficients $g_{00}$ and $g_{rr}$ are related to each other through the equation below ($u\equiv r^2/R^2$). Using it one can check whether the new metric satisfies it. $${\rm e}^{-\lambda}\frac{\rm d}{{\rm d} u} \left( {\rm e}^{-\lambda}\frac{\rm d}{{\rm d}u}{\rm e}^{\nu}\right) = \frac{1}{4}\frac{\rm d}{{\rm d}u}\left(\frac{1-{\rm e}^{-2\lambda}}{u}\right)~{\rm e}^{\nu}~.\tag{1}$$
The easiest way to do it is to notice that both metrics satisfy ${\rm e}^{\nu}={\rm e}^{-\lambda}$ what simplifies the equation $(1)$ to
\begin{equation}\label{compactform2}
\frac{\rm d}{{\rm d} u} \left(\frac{\rm d}{{\rm d}u}{\rm e}^{\nu}\right) = \frac{1}{2}\frac{\rm d}{{\rm d}u}\left(\frac{1-{\rm e}^{-2\lambda}}{u}\right),~\tag{2}
\end{equation}
The solution of equation $(2)$ reads $${\rm e}^{-2\lambda}=1+\frac{C_{1}}{\sqrt{u}}+C_{2}~u\equiv1+\frac{\tilde{C}_{1}}{r}+\tilde{C}_{2}~r^2.~\tag{3}$$
The vacuum solution (Schwarzschild solution) corresponds to $\tilde{C_{2}}\equiv0$.
Concluding, the only metric satisfying $~g_{00}\cdot g_{rr}=1$ relation is the Schwarzschild metric (with or without cosmological term).
It is possible to obtain a new metric of the form you have given, but it would not describe a black hole.
To derive this metric, we start with the original Schwarzchild metric: \begin{equation} \mathrm{d}s^2 = -\left(1-\dfrac{2GM}{r}\right)\mathrm{d}t^2+ \left(1-\dfrac{2GM}{r}\right)^{-1}\mathrm{d}r^2+ r^2\mathrm{d}\phi^2 \end{equation}
Now we can write this in matrix form as: \begin{equation} g_{\mu\nu}= \begin{bmatrix} -\left(1-\dfrac{2GM}{r}\right) & 0 & 0 \\ 0 & \left(1-\dfrac{2GM}{r}\right)^{-1} & 0 \\ 0 & 0 & \dfrac{r^2}{\left(1-\frac{2GM}{r}\right)} \end{bmatrix},\ \textrm{for }\mu,\nu=t,r,\phi \end{equation}
For the next step, we need to find the inverse metric $g^{\mu\nu}$. This is given by: \begin{equation} g^{\mu\nu}= \begin{bmatrix} -\left(1-\dfrac{2GM}{r}\right)^{-1} & 0 & 0 \\ 0 & \left(1-\dfrac{2GM}{r}\right) & 0 \\ 0 & 0 & \dfrac{\left(1-\frac{2GM}{r}\right)}{r^2} \end{bmatrix},\ \textrm{for }\mu,\nu=t,r,\phi \end{equation}
Now we can use the Christoffel symbols to calculate the Riemann curvature tensor. The non-zero Christoffel symbols are given by: \begin{align*} \Gamma^t_{tr} =\Gamma^r_{tt} &= \dfrac{GM}{r^2} \\ \Gamma^r_{rr} &= -\dfrac{GM}{r^2}\left(1 - \dfrac{2GM}{r}\right) \\ \Gamma^\phi_{\phi r} = \Gamma^r_{\phi \phi} &= \dfrac{1}{2}r\left(1 - \dfrac{2GM}{r}\right)^{-1} \\ \Gamma^r_{\phi \phi} &= -r\left(1 - \dfrac{2GM}{r}\right)^{-1} \end{align*}
Using the Riemann curvature tensor, we can calculate the Ricci tensor and scalar curvature. However, in this case, it is easier to use the fact that the Ricci tensor is proportional to the metic. In other word, we can write: \begin{equation} R_{\mu\nu}=\dfrac{1}{2}Rg_{\mu\nu} \end{equation}
Using the original metric $g_{\mu\nu}$, we find: \begin{align} R_{tt}=R_{\phi\phi}&=+\dfrac{2GM}{r^3} \\ R_{rr}&=-\dfrac{2GM}{r^3} \end{align}
Substituting these values into the new metric, we obtain: \begin{align} \mathrm{d}s^2&=\left(1+\dfrac{2GM}{r}\right)dt^2+\left(1-\dfrac{2GM}{r}\right)^{-1}dr^2+r^2d\phi^2 \end{align}
