I think you need to assume that $K$ has a smooth boundary and is strictly convex to ensure that $g$ and $h$ are differentiable outside $0$.

Anyway, I do not think that the result is true. Assume that $K$ is the unit ball associated to the $\ell^p$-norm with $p>1$. Then $K^*$ is the unit ball associated to the $\ell^q$-norm with $q>1$ such that $1/p+1/q=1$. The functions $g$ and $h$ are the $\ell^p$-norm and the $\ell^q$-norm, so everything can be computed explicitly.

For every $x \in \mathbb{R}^2$, set $$x^{p-1}:=(x_1^{p-1}\mathrm{sign(x_1)},x_2^{p-1}\mathrm{sign(x_2)}).$$ Then $\nabla (g^p)(x) = (px_1^{p-1}\mathrm{sign(x_1)},px_2^{p-1}\mathrm{sign(x_2)}) = px^{p-1}$. By the chain rule, if $x$ is ot $0$, $$\nabla g(x) = \frac{1}{p}(g^p(x))^{1/p-1}\nabla (g^p)(x) = \frac{x^{p-1}}{g(x)^{p-1}}$$ $$x \times \nabla g(x) = \frac{x_1x_2^{p-1}-x_2x_1^{p-1}}{g(x)^{p-1}}.$$

My impression is that $g(x)=h(y)$ does not imply $x \times \nabla g(x) = y \times \nabla h(y)$. A counterexample may be found by taking $x=(1,0)$ and $y=2^{-1/q}(1,1)$. I did not push the computations forward.

I think you need to assume that $K$ has a smooth boundary and is strictly convex to ensure that $g$ and $h$ are differentiable outside $0$.

Anyway, I do not think that the result is true. Assume that $K$ is the unit ball associated to the $\ell^p$-norm with $p>1$. Then $K^*$ is the unit ball associated to the $\ell^q$-norm with $q>1$ such that $1/p+1/q=1$. The functions $g$ and $h$ are the $\ell^p$-norm and the $\ell^q$-norm, so everything can be computed explicitly.

For every $x \in \mathbb{R}^2$, set $$x^{p-1}:=(|x_1|^{p-1}\mathrm{sign(x_1)},|x_2|^{p-1}\mathrm{sign(x_2)}).$$ Then $\nabla (g^p)(x) = (p|x_1|^{p-1}\mathrm{sign(x_1)},p|x_2|^{p-1}\mathrm{sign(x_2)}) = px^{p-1}$. By the chain rule, if $x \ne 0$, $$\nabla g(x) = \frac{1}{p}(g^p(x))^{1/p-1}\nabla (g^p)(x) = \frac{x^{p-1}}{g(x)^{p-1}}$$ $$x \times \nabla g(x) = \frac{x_1x_2^{p-1}-x_2x_1^{p-1}}{g(x)^{p-1}}.$$

My impression is that $g(x)=h(y)$ does not imply $x \times \nabla g(x) = y \times \nabla h(y)$. A counterexample may be found by taking $x=(1,0)$ and $y=2^{-1/q}(1,1)$. I did not push the computations forward.

I think you need to assume that $K$ have smooth boundary and is strictly convex to ensure that $g$ and $h$ are differentiable outside $0$.

Anyway, I do not think that the result is true. Assume that $K$ is the unit ball associated to the $\ell^p$-norm with $p>1$. Then $K^*$ is is the unit ball associated to the $\ell^q$-norm with $q>1$ such that $1/p+1/q=1$. The functions $g$ and $h$ are the $\ell^p$-norm and the $\ell^q$-norm, so every thing can be compute explicitly.

For every $x \in \mathbb{R}^2$, set $$x^{p-1}:=(x_1^{p-1}\mathrm{sign(x_1)},x_2^{p-1}\mathrm{sign(x_2)}).$$ Then $\nabla (g^p)(x) = (px_1^{p-1}\mathrm{sign(x_1)},px_2^{p-1}\mathrm{sign(x_2)}) = px^{p-1}$. By the chain rule, if $x$ is ot $0$, $$\nabla g(x) = \frac{1}{p}(g^p(x))^{1/p-1}\nabla (g^p)(x) = \frac{x^{p-1}}{g(x)^{p-1}}$$ $$x \times \nabla g(x) = \frac{x_1x_2^{p-1}-x_2x_1^{p-1}}{g(x)^{p-1}}.$$

My impression is that $g(x)=h(y)$ does not imply $x \times \nabla g(x) = y \times \nabla h(y)$. A counterexample may be found by taking $x=(1,0)$ and $y=2^{-1/q}(1,1)$. I did not push the computations forward.

I think you need to assume that $K$ has a smooth boundary and is strictly convex to ensure that $g$ and $h$ are differentiable outside $0$.

Anyway, I do not think that the result is true. Assume that $K$ is the unit ball associated to the $\ell^p$-norm with $p>1$. Then $K^*$ is the unit ball associated to the $\ell^q$-norm with $q>1$ such that $1/p+1/q=1$. The functions $g$ and $h$ are the $\ell^p$-norm and the $\ell^q$-norm, so everything can be computed explicitly.

For every $x \in \mathbb{R}^2$, set $$x^{p-1}:=(x_1^{p-1}\mathrm{sign(x_1)},x_2^{p-1}\mathrm{sign(x_2)}).$$ Then $\nabla (g^p)(x) = (px_1^{p-1}\mathrm{sign(x_1)},px_2^{p-1}\mathrm{sign(x_2)}) = px^{p-1}$. By the chain rule, if $x$ is ot $0$, $$\nabla g(x) = \frac{1}{p}(g^p(x))^{1/p-1}\nabla (g^p)(x) = \frac{x^{p-1}}{g(x)^{p-1}}$$ $$x \times \nabla g(x) = \frac{x_1x_2^{p-1}-x_2x_1^{p-1}}{g(x)^{p-1}}.$$

My impression is that $g(x)=h(y)$ does not imply $x \times \nabla g(x) = y \times \nabla h(y)$. A counterexample may be found by taking $x=(1,0)$ and $y=2^{-1/q}(1,1)$. I did not push the computations forward.