Go from: hat$(H{)}_{\text{T}\text{O}\text{T}}=\frac{-{}^2}{2m_1}gra{d}_1^2-\frac{{}^2}{2m_2}gra{d}_2^2+V(x_{1}dots{z}_2)$

To

hat$(H{)}_{\text{T}\text{O}\text{T}}=$hat$(H{)}_{CM}(xYZ)+$hat$(H{)}_{VR}(xyz)$

hat$(H{)}_{CM}=\frac{-{}^2}{2M}gra{d}_{CM}^2$

hat$(H{)}_{VR}=\frac{-{}^2}{2}gra{d}_{VR}^2+V(xyz)$

Given

$x=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ etc. for $Y$ and $Z$

$x=x_2-x_1$ etc. for $y$ and $z$

$M=m_1+m_2,=\frac{m_1{m}_2}{m_1+m_2}$

$gra{d}_{CM}^2=\frac{de{l}^2}{del{x}^2}+\frac{de{l}^2}{del{Y}^2}+\frac{de{l}^2}{del{Z}^2}$

$gra{d}_{VR}^2=\frac{de{l}^2}{del{x}^2}+\frac{de{l}^2}{del{y}^2}+\frac{de{l}^2}{del{z}^2}$

and we assume $V(x_{1}dots{z}_2)=V(xyz)$ since it contains no derivatives. We will use the chain rule to make this happen:

$\frac{\text{d}\text{e}\text{l}}{del{x}_1}=\frac{\text{d}\text{e}\text{l}\text{x}}{del{x}_1}\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{x}}+\frac{\text{d}\text{e}\text{l}\text{x}}{del{x}_1}\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{x}}$

evaluating $\frac{\text{d}\text{e}\text{l}\text{x}}{del{x}_1}$ and $\frac{\text{d}\text{e}\text{l}\text{x}}{del{x}_1}$ we have

$\frac{\text{d}\text{e}\text{l}}{del{x}_1}=\frac{m_1}{m_1+m_2}\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{x}}-\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{x}}$

and $\frac{de{l}^2}{del{x}_1^2}=(\frac{\text{d}\text{e}\text{l}}{del{x}_1}{)}^2,$ Be careful and collect terms and you will get the correct answer!