You are confused about how constants are treated in different derivative rules. Read on to understand how y should be treated in your two different scenarios:

Consider the first example you gave:

f(x) = x^2 + y^2

[1] Differentiate both side respective to \delta x:

\frac{\delta}{\delta x} f(x) = \frac{\delta}{\delta x}(x^2 + y^2)

[2] **Sum rule of derivative**: the derivative of a sum is equal to the sum of the derivatives

\frac{\delta}{\delta x} f(x) = \frac{\delta}{\delta x}x^2 + \frac{\delta}{\delta x}y^2

[3] For the first term \frac{\delta}{\delta x}x^2, apply the **Power Rule**

\frac{\delta}{\delta x} f(x) = 2x + \frac{\delta}{\delta x}y^2

[4] For the second term \frac{\delta}{\delta x}y^2, since y^2 is the constant when differentiate respective to x, its value is 0.

\frac{\delta}{\delta x} f(x) = 2x + 0

*Remember that \frac{\delta}{\delta x} C = 0 (which means the constant will never change when we change x). This is why people automatically “ignore” ***summation terms without x** when do differentiation **with respect to x**.

[5] Simplify:

\frac{\delta}{\delta x} f(x) = 2x

Now consider the second example you gave:

f(x) = 3x^2y^3

[1] Again, differentiate both side with respective to \delta x

\frac{\delta}{\delta x} f(x) = \frac{\delta}{\delta x}(3x^2y^3)

[2] Since y^3 is constant when differentiate respective to x, we can rewrite:

\frac{\delta}{\delta x} f(x) = \frac{\delta}{\delta x}(3y^2 \times x^2)

*I think you was confused that y^2 can be ignored here because it can be “ignored” above. This is not correct as the y^2 is part of the whole term 3x^2y^3 (which contains x). We should see that 3y^2 is in fact the whole constant.*

[3] Apply the **Constant Multiple Rule**

\frac{\delta}{\delta x} f(x) = 3y^2\times\frac{\delta}{\delta x}(x^2)

[4] From there apply the **Power Rule**:

\frac{\delta}{\delta x} f(x) = 3y^2 \times 2 \frac{\delta}{\delta x}x

[5] Simplify:

\frac{\delta}{\delta x} f(x) = 6y^2\frac{\delta}{\delta x}x