Is it true that $\nabla_{\partial_i f}\partial_j f = \partial_{ij} f\ $? Let $f:M\to \Bbb R^K$ be a smooth function, $\nabla$ be the Euclidean connection on $\Bbb R^K$.

Is it true that $\nabla_{\partial_i f}\partial_j f = \partial_{ij} f\ $?


Edit: Everything after this point could be nonsense. Please feel free to ignore it and just consider the main question.

What I tried is writing $\partial_i f = \frac{\partial f^k}{\partial x^i}  \frac{\partial }{\partial y^k}$ thus
$$\begin{align}
\nabla_{\partial_i f}\partial_j f &= \left( \frac{\partial f^l}{\partial x^i} \cdot  \frac{\partial }{\partial y^l} \frac{\partial f^k}{\partial x^j} \right)  \frac{\partial }{\partial y^k}
\end{align}$$
but I don't know how to calculate this part
$$
\frac{\partial }{\partial y^l} \frac{\partial f^k}{\partial x^j}.
$$
Perhaps I'm overlooking something silly. Could someone help me confirming/disproving the statement?
 A: taking this up with respect to the previous discussion here Seeing that the second fundamental form is the orthogonal component of the Laplacian, I think you may have some confusion. 
What you are asking for this question, that $\nabla_{\partial_i f}{\partial_jf}=\partial_i\partial_j f  $, is certainly not true in general. Recall that for general coordinates ${x^1\ldots x^m}$ on a manifold $M$, the second derivative does not have any intrinsic meaning! The point of the covariant derivative is that it is intrinsically defined. $\nabla \colon TM \times TM \to TM$. So it always takes two tangent vectors and spits out a new tangent vector. So the left hand side in your expression is perfectly covariant (i.e. it is a tangent vector) whereas the right hand side is not, it may well vanish completely in one coordinate system and not some other (something certainly not true of a tangent vector!)
Now, there is one specific case in which your identity holds, and that is when we deal with a flat ambient connection, which was the context of the previous question. 
To illustrate that, suppose that $M$ is a submanifold of Euclidean space $\mathbb{R}^n$, and we will denote the flat connection on Euclidean space by $\bar{\nabla}$. Now let $\{y^1, \ldots, y^n\}$ be Cartesian coordinates on $\mathbb{R}^n$. That $\bar{\nabla}$ is flat means $\bar{\nabla}_{\frac{\partial}{\partial y^\alpha}} \frac{\partial}{\partial y^\beta}=0$. Let $f\colon M \to \mathbb{R^n}$ be the embedding of $M$ and let $\{x^1 ,\ldots , x^m\}$ be some local coordinates on $M$. Now the $\frac{\partial f}{\partial x^i}$ form a local frame for the tangent space to $M$. Then by the rules of the connection
$$ \bar{\nabla}_{\frac{\partial f}{\partial x^i}} \frac{\partial f}{\partial x^j} = \bar{\nabla}_{\frac{\partial f^\alpha}{\partial x^i} \frac{\partial}{\partial y^\alpha}} ( \frac{\partial f^\beta}{\partial x^j} \frac{\partial}{\partial y^\beta} ) = \frac{\partial f^\alpha}{\partial x^i} \bar{\nabla}_{\frac{\partial}{\partial y^\alpha}} ( \frac{\partial f^\beta}{\partial x^j} \frac{\partial}{\partial y^\beta} ) \\
=\frac{\partial f^\alpha}{\partial x^i} (\frac{\partial^2 f^\beta}{\partial y^\alpha \partial x^j} \frac{\partial}{\partial y^\beta} + \frac{\partial f^\beta}{\partial x^j}\bar{\nabla}_{\frac{\partial}{\partial y^\alpha}} \frac{\partial}{\partial y^\beta}) = \frac{\partial f^\alpha}{\partial x^i} \frac{\partial^2 f^\beta}{\partial y^\alpha \partial x^j} \frac{\partial}{\partial y^\beta} \\
=\frac{\partial^2 f^\beta}{\partial x^i \partial x^j} \frac{\partial}{\partial y^\beta} = \frac{\partial^2 f}{\partial x^i \partial x^j}  $$
where in moving to the last line we applied the chain rule.
