If $u={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$ , then by Euler’s theorem the value of $x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=$
1)$\sin u$
2)$\tan u$
3)$0$
4)$\cos u$

Hint: We will use Euler’s Homogeneous Theorem , which states that if $f$ is a homogeneous function of degree $n$ of variables $x,y,z$ then $x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}+z\dfrac{\partial f}{\partial z}=nf$ . We will first find the degree of given homogeneous function . Then we will apply the Euler’s Homogeneous Theorem and we will get our required answer.

Complete step-by-step solution:
A homogeneous function is a real valued function .
First we will learn how to check the given function is a homogeneous function or not
Let us give a function $f$ of two variables that is \[x\] & $y$ .
Now we will change $x$ into $tx$and $y$ into$ty$ .
So, our function will change from $f\left( x,y \right)$ to $f\left( tx,ty \right)$
Now we will try to take the maximum positive natural power of $t$ common from the function and we will represent the function as ${{t}^{n}}f\left( x,y \right)$ .
If we are able to write the function in this above form then we can say that our function is a homogeneous function .
$n$ is called a degree of homogeneity or we can say that $n$ represents the degree of homogeneous function and $n$ should be a real number only .
Euler’s Homogeneous Theorem states that if $f\left( x.y,z \right)$ is a homogeneous function then
$x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}+z\dfrac{\partial f}{\partial z}=nf$
Where $n$ is the degree of homogeneous function .
For this question we will use Euler’s Homogeneous Theorem of two variables only ,
If $f\left( x,y \right)$ is our given two variable function then $x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}=nf$where $n$ is the degree of given homogeneous function.
Given Function is
$u\left( x.y \right)={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$
First we will check whether the given function is homogeneous or not.
We will change the given function $u\left( x,y \right)$ into $u\left( tx,ty \right)$ .
$x$ Changes into \[tx\]$\And $ $y$ changes into $ty$
So, our function becomes
$\begin{align}
  & \\
 & u\left( tx,ty \right)={{\tan }^{-1}}\left( \dfrac{ty}{tx} \right) \\
\end{align}$
We can also write it as,
$u\left( tx,ty \right)={{t}^{0}}{{\tan }^{-1}}\left( \dfrac{y}{x} \right)$
We can see that power of $t$ is $0$
So we can conclude that, the degree of given homogeneous function is $0$ that is$n=0$
Then we will substitute the value of $n$ in above stated Euler’s Homogeneous Theorem of two variables
$x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=0\times {{\tan }^{-1}}\left( \dfrac{y}{x} \right)$
$\therefore x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=0$
$\therefore $So the required answer is $0$
Hence , The Correct Option is $3$ .

Note:Homogenous word is used for real valued functions. We will see the word homogeneous in other mathematics fields also like in the system of linear equations , differential equations. But in different topics it will represent different meanings .Homogeneous equations can be solved in a particular format.