This question is about a mistake in working. Consider the gravitational two-body problem, where two particles $\mathbf{r}_1, \mathbf{r}_2 $ with masses $m_1, m_2$ are attracted to each other under the inverse square law. So the potential is $$U=- \frac{Gm_1m_2}{|{\mathbf{r}_1-\mathbf{r}_2}|} .$$
The force on a body is equal to the negative gradient of $U$ with respect to its coordinates. Then: \begin{align}-\frac{\partial U}{\partial \mathbf{r}_1} &= \frac{\partial}{\partial \mathbf{r}_1} \frac{Gm_1m_2}{|{\mathbf{r}_1-\mathbf{r}_2}|} \\ & = -\frac{Gm_1m_2}{2[({\mathbf{r}_1-\mathbf{r}_2}) \cdot ({\mathbf{r}_1-\mathbf{r}_2})]^\frac{3}{2}}(({\mathbf{r}_1-\mathbf{r}_2})\cdot \mathbf{1})+\mathbf{1} \cdot ({\mathbf{r}_1-\mathbf{r}_2}))\\ &= \frac{Gm_1m_2}{|{\mathbf{r}_1-\mathbf{r}_2}|^3 }(({\mathbf{r}_1-\mathbf{r}_2}) \cdot \mathbf{1}). \end{align}
However, the force must be a vector - which step is incorrect?
$\frac{\partial}{\partial\mathbf x}(\mathbf a\cdot\mathbf x)$ is $\mathbf a$, not $\mathbf a\cdot\mathbf 1$. In coordinates:
$$ \pmatrix{\frac{\partial}{\partial x}\\\frac{\partial}{\partial y}\\\frac{\partial}{\partial z}}(a_xx+a_yy+a_zz)=\pmatrix{a_x\\a_y\\a_z}\;. $$