In an isothermal process, how can the change in internal energy be 0? It was written in my textbook,

$$
\mathrm{d}U = 
  \left(\frac{\partial U}{\partial T}\right)_V \mathrm{d}T 
 +\left(\frac{\partial U}{\partial V}\right)_T \mathrm{d}V
$$
If the process is isothermal, $\mathrm{d}T = 0$. So, the equation reduces to:

$$
\mathrm{d}U = \left(\frac{\partial U}{\partial V}\right)_T \mathrm{d}V
$$

I was told, $\mathrm{d}U = 0$ in an isothermal process, does that mean $\mathrm{d}V = 0$, or  $\left(\frac{\partial U}{\partial V}\right) = 0$? But how is it possible that $\mathrm{d}V=0$ in an isothermal process, or $\left(\frac{\partial U}{\partial V}\right) = 0$?

If someone could explain what am I doing wrong, that would be very helpful.
The internal energy of an ideal gas is simply $$ U = \alpha nRT,$$
where $$\alpha = \frac{\text{degrees of freedom}}{2}$$
So, in an isothermal process,
$$
\Delta T = 0 \Longrightarrow \Delta U = \alpha nR\Delta T = 0,$$
and likewise any
$$\left(\frac{\partial U}{\partial P}\right)_T = \left(\frac{\partial U}{\partial V}\right)_T = 0.$$