[tex]\csc y=-\dfrac{10}3\implies\sin y=-\dfrac3{10}[/tex]
[tex]\cos^2y=1-\sin^2y\implies \cos y=\pm\sqrt{1-\dfrac9{100}}=\pm\dfrac{\sqrt{91}}{10}[/tex]
You're told that [tex]\cos y<0[/tex], so omit the positive root.
[tex]\cot y=\dfrac{\cos y}{\sin y}=\dfrac{-\frac{\sqrt{91}}{10}}{-\frac3{10}}=\dfrac{\sqrt{91}}3[/tex]
