Step 1: Recall what "linear" means.
A differential equation is linear in $y$ if it can be written as $\frac{dy}{dx} + P(x)\,y = Q(x)$, with no powers or products of $y$ and its derivative. The same idea, with roles swapped, defines linear in $x$.
Step 2: Rearrange the first equation.
From $\frac{dy}{dx}(x+y+1)=1$ we get $\frac{dx}{dy} = x + y + 1$.
Step 3: Test the first for linearity in $x$.
Rewrite as $\frac{dx}{dy} - x = y + 1$. This is exactly the linear form with $x$ as the dependent variable and $y$ as the independent one. So equation one is linear in $x$.
Step 4: Look at the second equation.
The second is $\frac{dx}{dy} = 3y + 2x^2$. Treated as an equation for $x$, the $x^2$ term breaks linearity in $x$.
Step 5: View the second from the $y$ side.
Reading it for $y$ instead, the right side is linear in $y$ (the $3y$ term is first power, and $2x^2$ acts as a known function). So this equation behaves as linear in $y$.
Step 6: Combine the findings.
Equation one is linear in $x$, and equation two is linear in $y$. That matches the option below.
\[ \boxed{\text{One is linear in } x \text{ and the other is linear in } y} \]