In the book The finite element method - Theory, implementation and applications by Larson et al. there is a fairly general procedure for finding the shape functions. It is based on the Vandermonde matrix and goes as follows:
A general second-order polynomial $v(x_1,x_2,x_3)$ on the tetrahedron is of the form
$$v(x_1,x_2,x_3) = c_0 + c_1 x_1 + c_2 x_2 + c_3 x_3 + c_4 x_1 x_2 + c_5 x_2 x_3 + c_6 x_1 x_3 + c_7 x_1^2 + c_8 x_2^2 + c_9 x_3^2.$$
The degrees of freedom $L_j$, $j\in\{1,\cdots\!,10\}$, are the nodal values and the edge middle point values:
$$L_1(v) = v(0,0,0);\quad L_2(v)=v(1,0,0);\quad \cdots \quad L_{10}(v)=v(0,0.5,0.5).$$
Note that you may choose the order as you wish. Then the Vandermonde matrix is
$$\begin{pmatrix} L_1(1) & L_1(x_1) & L_1(x_2) & \cdots & L_1(x_3^2) \\
L_2(1) & L_2(x_1) & L_2(x_2) & \cdots & L_2(x_3^2) \\
L_3(1) & L_3(x_1) & L_3(x_2) & \cdots & L_3(x_3^2) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
L_{10}(1) & L_{10}(x_1) & L_{10}(x_2) & \cdots & L_{10}(x_3^2) \end{pmatrix}.$$
By evaluating the elements of this matrix and taking its inverse you may read the respective coefficients $c_0,\cdots,c_9$ from the columns of the inverse matrix. For this particular case I made a short Mathematica-script to evaluate the coefficients and it revealed that the shape functions are
$$\begin{aligned}
\phi_1&=1-3x_1-3x_2-3x_3+4x_1 x_2 + 4 x_2 x_3 + 4 x_1 x_3+2x_1^2+2x_2^2+2x_3^2,\\
\phi_2&=-x_1+2x_1^2,\\
\phi_3&=-x_2+2x_2^2,\\
\phi_4&=-x_3+2x_3^2,\\
\phi_5&=4x_1-4x_1x_2-4x_1x_3-4x_1^2,\\
\phi_6&=4x_2-4x_1x_2-4x_2x_3-4x_2^2,\\
\phi_7&=4x_3-4x_2x_3-4x_1x_3-4x_3^2,\\
\phi_8&=4x_1x_2,\\
\phi_8&=4x_1x_3,\\
\phi_9&=4x_2x_3.\\
\end{aligned}
$$
I did not verify the shape functions in any way. Thus, you may want to redo the calculations.