For i), is the answer $df/du = (a)df/dx + (2cu)df/dy, df/dv = (b)df/dx + (2dv)df/dy$
For ii), is the answer $d^2 f/du^2 = (a^2) d^2 f/dx^2 + (4acu) d^2 f/dxdy + (4c^2 u^2) d^2 f/dy^2$ ,
$d^2 f/dv^2 = (b^2) d^2 f/dx^2 + (4bdv) d^2 f/dxdy + (4d^2 v^2) d^2 f/dy^2$
$x(u,v)=au+bv$ $\quad$ $y(u,v)=cu^2+dv^2$
So $\dfrac{\partial x}{\partial u}=a$ and $\dfrac{\partial x}{\partial v}=b$
and $\dfrac{\partial y}{\partial u}=2cu$ and $\dfrac{\partial y}{\partial v}=2dv$
Now use the chain rule :
(i)
$\therefore \dfrac{\partial f}{\partial u}=\dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial u}=a\dfrac{\partial f}{\partial x}+2cu\dfrac{\partial f}{\partial y}$.
$\dfrac{\partial f}{\partial v}=\dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial v}=b\dfrac{\partial f}{\partial x}+2du\dfrac{\partial f}{\partial y}$.
From here I think you can proceed. For further help with partial derivatives check this out !!