= Solution
Apply the <product rule> and the <chain rule> once more to each expression in part (i). The <second derivative chain rule> gives
$$
\boxed{f_{xx}=2f_u+2y^2f_v+4x^2f_{uu}+8x^2y^2f_{uv}+4x^2y^4f_{vv},}
$$
$$
\boxed{f_{yy}=2f_u+2x^2f_v+4y^2f_{uu}+8x^2y^2f_{uv}+4x^4y^2f_{vv},}
$$
and
$$
\boxed{f_{xy}=4xyf_{uu}+4xy(x^2+y^2)f_{uv}+4x^3y^3f_{vv}+4xyf_v.}
$$
For example, the coefficient of $f_{uv}$ in the mixed <partial derivative> is $u_xv_y+v_xu_y=4x^3y+4xy^3$, while the final term is $v_{xy}f_v=4xyf_v$. The symmetry $f_{uv}=f_{vu}$ is used in combining the mixed terms under the usual twice-continuously-differentiable interpretation.
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