The likelihood function is

\begin{array}{l} L (\lambda) = f (y _ {1}) \cdot f (y _ {2}) \cdot \dots \cdot f (y _ {n}) = \left[ \sqrt {\frac {2}{\pi}} \cdot \frac {y _ {1} ^ {2}}{\lambda^ {3}} \cdot e ^ {- \frac {y _ {1} ^ {2}}{2 \lambda^ {2}}} \right] \cdot \dots \cdot \left[ \sqrt {\frac {2}{\pi}} \cdot \frac {y _ {n} ^ {2}}{\lambda^ {3}} \cdot e ^ {- \frac {y _ {n} ^ {2}}{2 \lambda^ {2}}} \right] \\ = \left(\sqrt {\frac {2}{\pi}}\right) ^ {n} \cdot \frac {y _ {1} ^ {2} \ldots y _ {n} ^ {2}}{\lambda^ {3 n}} \cdot e ^ {- \frac {y _ {1} ^ {2} + \ldots + y _ {n} ^ {2}}{2 \lambda^ {2}}}, \\ \end{array}

and the log likelihood function is

$\ln L(\lambda)=\ln\left(\sqrt{\frac{2}{\pi}}\right)^{n}+\ln(y_{1}^{2}\ldots y_{n}^{2})-3n\ln\lambda-\frac{y_{1}^{2}+\ldots+y_{n}^{2}}{2\lambda^{2}}.$

Now taking derivative and setting it as zero give

$0=\ln^{\prime}L(\lambda)=-\frac{3n}{\lambda}+\frac{y_{1}^{2}+\ldots+y_{n}^{2}}{\lambda^{3}}\Longrightarrow\hat{\lambda}_{ML}=\sqrt{\frac{y_{1}^{2}+\ldots+y_{n}^{2}}{3n}}.$