I'd like to show that the Newton-Leibniz rule works for Stratonovich integrals for multidimensions. I use $\circ dW_t$ to denote the Stratonovich differential. We know that if $X_t$ is the solution to the Stratonovich sense SDE $$dX_t=b(X_t,t)dt+\sigma(X_t,t)\circ dW_t$$, then it is also the solution of the Ito sense SDE $$dX_t=\left(b(X_t,t)+\frac{1}{2}\partial_x\sigma\sigma(X_t,t)dt\right)dt+\sigma(X_t,t)dW_t$$
For the multidimensional case, here is my proof: if $X_t$ is the solution to the Stratonovich sense SDE $$dX_t=b(X_t,t)dt+\sigma(X_t,t)\circ dW_t$$ then it satisfies the Ito sense SDE $$dX_t=\left(b(X_t,t)+\frac{1}{2}\nabla_x\sigma:\sigma(X_t,t)\right)dt+\sigma(X_t,t)dW_t$$ where $(\nabla_x\sigma:\sigma)_i=\sum_{j,k}\partial_k\sigma_{ij}\sigma_{kj}$. By multidimensional Ito formula, $$df(X_t)=\left(b\cdot\nabla f+\frac{1}{2}\sigma\sigma^T\nabla^2f\right)dt+\nabla f\sigma(X_t,t)dW_t=\nabla f(X_t)\cdot b(X_t,t)dt+\frac{1}{2}\sigma\sigma^T\nabla^2f(X_t)dt+\nabla f(X_t)\cdot\sigma\cdot dW_t$$
I know that the final answer should be $$\nabla f(X_t)\cdot b(X_t,t)dt+\nabla f(X_t)\cdot\sigma(X_t,t)\circ dW_t$$ but can I equate $$\frac{1}{2}\sigma\sigma^T\nabla^2f(X_t)dt+\nabla f(X_t)\cdot\sigma\cdot dW_t\text{ and }\nabla f(X_t)\cdot\sigma(X_t,t)\circ dW_t$$ by the above reasoning? Personally, I think this should be justified more carefully. Can anyone provide any answers regarding this?
Edit: Here is my new method to solve this: $$f(X_t)-f(X_0)=\int_0^tf'(X_s)\circ dX_s=\int_0^t\mathcal{L}f(X_s)ds+\int_0^tf'(X_s)\sigma dW_s =\int_0^t \left[f'\left(\frac{b\sigma\sigma'}{2}\right)+\frac{f''\sigma^2}{2}\right]ds+\int_0^t f'\sigma dW_s=\int_0^t\left[f'b+\frac{\sigma}{2}(f'\sigma)'\right]ds+\int_0^tf'\sigma dW_s$$
Does the terms on the right-hand side correspond the multi-dimensional Ito formula? I'd like to get back to $$df(X_t)=\nabla f(X_t)\cdot b(X_t,t)dt+\nabla f(X_t)\cdot\sigma(X_t,t)\circ dW_t$$ , so how can I proceed next?