Answer
$\dfrac{\partial K}{\partial m}\dfrac{\partial^{2}K}{\partial v^{2}}=K$
Work Step by Step
For $ \dfrac{\partial K}{\partial m}$, here all variables aside from $m$ to be treated as constants.
$ \dfrac{\partial K}{\partial m}=\dfrac{\partial}{\partial m}[\dfrac{1}{2}mv^{2}]=\dfrac{1}{2}v^{2}$
$\dfrac{\partial K}{\partial v}=\dfrac{\partial}{\partial v}[(\dfrac{1}{2})mv^{2}]=mv$
This gives: $ \dfrac{\partial^{2}K}{\partial v^{2}}=\dfrac{\partial}{\partial v}[mv]=m$
and $\dfrac{\partial K}{\partial m}\times\dfrac{\partial^{2}K}{\partial v^{2}}=\dfrac{1}{2}v^{2}m=K$
Hence, we have $\dfrac{\partial K}{\partial m}\dfrac{\partial^{2}K}{\partial v^{2}}=K$