How to prove that the expression $\varphi_{,ij}:=\frac{\partial^2\varphi}{\partial x_i\partial x_j}=\nabla\nabla\varphi$ is a tensor of second order where $\varphi$ is a scalar? Furthermore, how to prove that $a\times b:=a_i b_j\varepsilon_{ijk}e_k$ is a vector?
We can either prove it by definition or use the so-called "tensor recognition theorem" claiming that if $p_{i_1i_2\cdots i_mj_1j_2\cdots j_n}q_{j_1\cdots j_n} = r_{i_1\cdots i_m}$, then $p$ must be a tensor of order $m+n$, where $q_{j_1\cdots j_n}$ is a tensor of order $n$ and $r_{i_1\cdots i_m}$ a tensor of order $m$.
$1$) Suppose that $O X_1 X_2 X_3$ is the coordinate system corresponding to the given basis $e_1, e_2, e_3$ on $E_3$. The gradient of a scalar function $\varphi (X_1,X_2,X_3)$, using this coordinate system, is defined by $$\nabla \phi(X_1,X_2,X_3) = \frac{\partial \varphi}{\partial X_i} e_i.$$
Suppose that $\tilde{A} = \{ \tilde{e}_1,\tilde{e}_2,\tilde{e}_3 \}$ is another orthonormal basis with corresponding cartesian coordinates $O \tilde{X_1} \tilde{X_2} \tilde{X_3}$ given by $\tilde{X}_i = q_{ij} X_j$, where $Q = (q_{ij}) \in SO(3)$.
By the chain rule, $$\frac{\partial \tilde{\varphi}}{\partial \tilde{X}_i} = \frac{\partial \varphi}{\partial X_k} \frac{\partial X_k}{\partial \tilde{X}_i}.$$
Using $\tilde{X}_i = q_{ij} X_j$, we have $$\frac{\partial X_k}{\partial \tilde{X}_i} = q_{ik},$$ and hence $$\frac{\partial \tilde{\varphi}}{\partial \tilde{X}_i} = q_{ik} \frac{\partial \varphi}{\partial X_k}.$$
That is, by definition, $\nabla \varphi$ is a cartesian tensor of order $1$. Doing this again, but with $\nabla \varphi$ in place of $\varphi$, shows that ${\nabla}^{2} \varphi$ is a cartesian tensor of order $2$.
$2$) $a$ and $b$ are vectors so $a_i$ and $b_j$ are components of cartesian tensors of order $1$, and the alternating tensor $\varepsilon = \varepsilon_{klm}$ is a cartesian tensor of order $3$.
Taking their product we get that $a_i b_j \varepsilon_{klm}$ are the components of a cartesian tensor of order $5$.
Contracting indices (that is, setting two indices equal and thus effecting a sum) gives that $a_i b_j \varepsilon_{ilm}$ are the components of a cartesian tensor of order $4$, and contracting again gives that $a_i b_j \varepsilon_{ijm}$ are the components of a cartesian tensor of order $3$.
Hence, $a \times b$ is a cartesian tensor of order $3$.