Answer
\[{y^{,,}} = \frac{{4{e^{2y}}}}{{\,\left( {1 - 2{e^{2y}}} \right)^3}}\]
Work Step by Step
\[\begin{gathered}
{e^{2y}} + x = y \hfill \\
\hfill \\
use\,\,the\,\,Implicit\,\,differentiation \hfill \\
\hfill \\
2{y^,}\,{e^{2y}} + 1 = {y^,} \hfill \\
\hfill \\
factor\,\,{y^,} \hfill \\
\hfill \\
{y^,} = \,\left( {2{e^{2y}} - 1} \right) = - 1 \hfill \\
\hfill \\
solve\,\,for\,\,{y^,} \hfill \\
\hfill \\
{y^,} = \frac{1}{{1 - 2{e^{2y}}}} \hfill \\
\hfill \\
Implicit\,\,differentiation \hfill \\
\hfill \\
{y^{,,}} = \frac{{ - 4{y^,}{e^{2y}}}}{{\,{{\left( {1 - 2{e^{2y}}} \right)}^2}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
{y^{,,}} = \frac{{4\,\left( {\frac{1}{{1 - 2{e^{2y}}}}} \right){e^{2y}}}}{{\,\left( {1 - 2{e^{2y}}} \right)^2}} \hfill \\
\hfill \\
{y^{,,}} = \frac{{4{e^{2y}}}}{{\,\left( {1 - 2{e^{2y}}} \right)^3}} \hfill \\
\hfill \\
\end{gathered} \]
