How do you differentiate x e x 2 + y 2

To differentiate a two-variable function, you need to build the gradient. The gradient is a vector with as many coordinates as the variables the function depends on.
Each coordinate of the vector is a derivative with respect to one of the variables. So, in the two-variables case, you need to calculate the derivatives with respect to x and y, and then put them together in a vector.
Since deriving with respect to a variable means to consider the other as a constant, it's easier to derivate your function if it's expressed in the form
${\displaystyle x{e}^{x^2+y^2}=x\cdot e^{x^2}\cdot e^{y^2}}$
So, deriving with respect to x, and using the product rule ${\displaystyle \left(fg\right)\prime =f\prime g+fg\prime }$ where ${\displaystyle f\left(x\right)=x}$ and ${\displaystyle g\left(x\right)=e^{x^2}}$, we get
${\displaystyle \frac{d}{dx}x\cdot e^{x^2}\cdot e^{y^2}=e^{y^2}(e^{x^2}+x\cdot e^{x^2}\cdot 2x)=}$
${\displaystyle e^{x^2}\cdot e^{y^2}(1+2x^2)=e^{x^2+y^2}(1+2x^2)}$
Where the derivative of ${\displaystyle e^{x^2}}$ has been calculated using the chain rule, which states that ${\displaystyle f\left(g\left(x\right)\right)\prime =f\prime \left(g\left(x\right)\right)\cdot g\prime \left(x\right)}$, where ${\displaystyle f\left(x\right)=e^x}$, and ${\displaystyle g\left(x\right)=x^2}$
The derivative with respect to y is easier, since the only factor to differentiate is ${\displaystyle e^{y^2}}$, while the others depend only on x and are thus to be consideblack as constant. So, we have
${\displaystyle \frac{d}{dy}x\cdot e^{x^2}\cdot e^{y^2}=x\cdot e^{x^2}\cdot e^{y^2}\cdot 2y=}$
${\displaystyle 2xy{e}^{x^2+y^2}}$
The gradient is thus the vector
${\displaystyle (e^{x^2+y^2}(1+2x^2),2xy e^{x^2+y^2})}$