\begin {align*} y^{\prime \prime } & =-\frac {1}{2\left ( y^{\prime }\right ) ^{2}}\\ 2\left ( y^{\prime }\right ) ^{2}y^{\prime \prime } & =-1 \end {align*}
With IC \begin {align*} y\left ( 0\right ) & =1\\ y^{\prime }\left ( 0\right ) & =-1 \end {align*}
Integrating both sides gives\begin {align*} \int 2\left ( y^{\prime }\right ) ^{2}y^{\prime \prime }dx & =\int -dx\\ \frac {2}{3}\left ( y^{\prime }\right ) ^{3} & =-x+c\\ \left ( y^{\prime }\right ) ^{3} & =-\frac {3}{2}x+c_{1} \end {align*}
Hence \begin {align} y_{1}^{\prime } & =\left ( -\frac {3}{2}x+c_{1}\right ) ^{\frac {1}{3}}\tag {1}\\ y_{2}^{\prime } & =-\left ( -1\right ) ^{\frac {1}{3}}\left ( -\frac {3}{2}x+c_{1}\right ) ^{\frac {1}{3}}\tag {2}\\ y_{3}^{\prime } & =\left ( -1\right ) ^{\frac {2}{3}}\left ( -\frac {3}{2}x+c_{1}\right ) ^{\frac {1}{3}} \tag {3} \end {align}
Trying solution (1). Integrating gives\begin {align*} y_{1} & =\int \left ( -\frac {3}{2}x+c_{1}\right ) ^{\frac {1}{3}}dx+c_{2}\\ & =-\frac {1}{2}\left ( -\frac {3}{2}x+c_{1}\right ) ^{\frac {4}{3}}+c_{2} \end {align*}
Applying \(y\left ( 0\right ) =1\) gives\begin {equation} 1=-\frac {1}{2}c_{1}^{\frac {4}{3}}+c_{2} \tag {4} \end {equation} And \(y^{\prime }\left ( x\right ) \) gives\[ y_{1}^{\prime }=\left ( -\frac {3}{2}x+c_{1}\right ) ^{\frac {1}{3}}\] Hence \(y^{\prime }\left ( 0\right ) =-1\) gives\[ -1=c_{1}^{\frac {1}{3}}\] No solution. Trying solution (2). Integrating gives\begin {align} y_{2} & =-\left ( -1\right ) ^{\frac {1}{3}}\int \left ( -\frac {3}{2}x+c_{1}\right ) ^{\frac {1}{3}}dx+c_{2}\nonumber \\ & =-\left ( -1\right ) ^{\frac {1}{3}}\left ( -\frac {1}{2}\left ( -\frac {3}{2}x+c_{1}\right ) ^{\frac {4}{3}}\right ) +c_{2} \tag {4A} \end {align}
Applying \(y\left ( 0\right ) =1\) gives\begin {equation} 1=\left ( -1\right ) ^{\frac {1}{3}}\left ( \frac {1}{2}c_{1}^{\frac {4}{3}}\right ) +c_{2} \tag {5} \end {equation} And \(y_{2}^{\prime }\left ( x\right ) \) gives\[ y_{2}^{\prime }\left ( x\right ) =-\left ( -\frac {1}{2}\right ) ^{\frac {1}{3}}\left ( -3x+2c_{1}\right ) ^{\frac {1}{3}}\] Hence \(y^{\prime }\left ( 0\right ) =-1\) gives\begin {align*} -1 & =-\left ( -\frac {1}{2}\right ) ^{\frac {1}{3}}\left ( 2c_{1}\right ) ^{\frac {1}{3}}\\ 1 & =\left ( -1\right ) ^{\frac {1}{3}}\left ( c_{1}\right ) ^{\frac {1}{3}} \end {align*}
No solution. Finally we will try \(y_{3}.\) Integrating gives\begin {align*} y_{3} & =\left ( -1\right ) ^{\frac {2}{3}}\int \left ( -\frac {3}{2}x+c_{1}\right ) ^{\frac {1}{3}}+c_{2}\\ & =\left ( -1\right ) ^{\frac {2}{3}}\left ( -\frac {1}{2}\left ( -\frac {3}{2}x+c_{1}\right ) ^{\frac {4}{3}}\right ) +c_{2} \end {align*}
Applying \(y\left ( 0\right ) =1\) gives\begin {equation} 1=\left ( -1\right ) ^{\frac {2}{3}}\left ( -\frac {1}{2}c_{1}^{\frac {4}{3}}\right ) +c_{2} \tag {6} \end {equation} And \(y_{3}^{\prime }\left ( x\right ) \) gives\[ y_{3}^{\prime }\left ( x\right ) =\left ( -1\right ) ^{\frac {2}{3}}\left ( -\frac {3}{2}x+c_{1}\right ) ^{\frac {1}{3}}\] Hence \(y^{\prime }\left ( 0\right ) =-1\) gives\[ -1=\left ( -1\right ) ^{\frac {2}{3}}\left ( c_{1}\right ) ^{\frac {1}{3}}\] Solving gives \(c_{1}=-1\). Substituting into (6) gives\begin {align*} 1 & =\left ( -1\right ) ^{\frac {2}{3}}\left ( -\frac {1}{2}\left ( -1\right ) ^{\frac {4}{3}}\right ) +c_{2}\\ c_{2} & =\frac {3}{2} \end {align*}
Hence solution is\begin {align*} y_{3} & =\left ( -1\right ) ^{\frac {2}{3}}\left ( -\frac {1}{2}\left ( -\frac {3}{2}x+c_{1}\right ) ^{\frac {4}{3}}\right ) +c_{2}\\ & =\left ( -1\right ) ^{\frac {2}{3}}\left ( -\frac {1}{2}\left ( -\frac {3}{2}x-1\right ) ^{\frac {4}{3}}\right ) +\frac {3}{2}\\ & =\frac {3}{2}-\frac {1}{2}\left ( -1\right ) ^{\frac {3}{2}}\left ( -\frac {3}{2}x-1\right ) ^{\frac {4}{3}} \end {align*}
This problem shows that out of the 3 solutions, only one was valid.