I've been working the following problem:
If $T_i^j$ is a type (1,1) tensor field show that $$ H_{ij}^k = T_i^r \dfrac{\partial T_j^k}{\partial x^r} - T_j^r \dfrac{\partial T_i^k}{\partial x^r} + T_r^k \left( \dfrac{\partial T_i^r}{ \partial x^j} - \dfrac{\partial T_j^r}{\partial x^i} \right) $$ is a type (1,2) tensor.
$-$From Tensors, Differential Forms, and Variational Principles by Lovelock and Rund (p.99).
and after introducing the transformation $x^i = x^i(\bar{x}^\alpha)$ and expanding, I'm able to show $$ \begin{align*} H_{ij}^k &= \dfrac{\partial \bar{x}^\alpha}{\partial x^i} \dfrac{\partial \bar{x}^\beta}{\partial x^j} \dfrac{\partial x^k}{\partial \bar{x}^\gamma} \overline{H}_{\alpha \beta}^\gamma \\ &+ \dfrac{\partial \bar{x}^\mu}{\partial x^i} \dfrac{\partial \bar{x}^\nu}{\partial x^j} \dfrac{\partial x^k}{\partial \bar{x}^\lambda} \dfrac{\partial \bar{x}^\rho}{\partial x^r} \dfrac{\partial^2 x^r}{\partial \bar{x}^\sigma \partial \bar{x}^\nu} \overline{T}_\rho^\lambda \overline{T}_\mu^\sigma - \dfrac{\partial \bar{x}^\mu}{\partial x^j} \dfrac{\partial \bar{x}^\nu}{\partial x^i} \dfrac{\partial x^k}{\partial \bar{x}^\lambda} \dfrac{\partial \bar{x}^\rho}{\partial x^r} \dfrac{\partial^2 x^r}{\partial \bar{x}^\sigma \partial \bar{x}^\nu} \overline{T}_\rho^\lambda \overline{T}_\mu^\sigma \end{align*} $$ by noting that $$\begin{align*} T_a^b &= \dfrac{\partial \bar{x}^\mu}{\partial x^a} \dfrac{\partial x^b}{\partial \bar{x}^\nu}\overline{T}_\mu^\nu \\ \dfrac{\partial T_a^b}{\partial x^c} &= \dfrac{\partial \bar{x}^\lambda}{\partial x^c} \dfrac{\partial \bar{x}^\mu}{\partial x^a} \dfrac{\partial x^b}{\partial \bar{x}^\nu} \dfrac{\partial \overline{T}_\mu^\nu}{\partial \bar{x}^\lambda} + \dfrac{\partial \bar{x}^\lambda}{\partial x^c} \dfrac{\partial \bar{x}^\mu}{\partial x^a} \dfrac{\partial^2 x^b}{\partial \bar{x}^\nu \partial \bar{x}^\lambda} \overline{T}_\mu^\nu, \end{align*}$$
but I haven't been able to get the last two terms to cancel. Any help would be appreciated.
As Ted mentioned in the comments, my derivative of $T$ was missing a term: $$\dfrac{\partial T_a^b}{\partial x^c} = \dfrac{\partial \bar{x}^\lambda}{\partial x^c} \dfrac{\partial \bar{x}^\mu}{\partial x^a} \dfrac{\partial x^b}{\partial \bar{x}^\nu} \dfrac{\partial \overline{T}_\mu^\nu}{\partial \bar{x}^\lambda} + \dfrac{\partial \bar{x}^\lambda}{\partial x^c} \dfrac{\partial \bar{x}^\mu}{\partial x^a} \dfrac{\partial^2 x^b}{\partial \bar{x}^\nu \partial \bar{x}^\lambda} \overline{T}_\mu^\nu + \dfrac{\partial x^b}{\partial \bar{x}^\nu} \dfrac{\partial \bar{x}^\mu}{\partial x^a \partial x^c} \overline{T}_\mu^\nu. $$ After correcting this term and using identity (3.19) from Lovelock and Rund, I am able to show that $$ H_{ij}^k = \dfrac{\partial \bar{x}^\alpha}{\partial x^i} \dfrac{\partial \bar{x}^\beta}{\partial x^j} \dfrac{\partial x^k}{\partial \bar{x}^\gamma} \overline{H}_{\alpha \beta}^\gamma $$ as desired.