I need to calculate $\nabla f$ and $\nabla f(0,0)$, where $$f(x,y)=10x^3-5x^2+5xy+5y^2+8.$$
My work so far: Using the formula
$$\nabla f = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$$ I got \begin{align*} \frac{\partial }{\partial x}(10x^3-5x^2+5xy+5y^2+8)&=30 x^{2} - 10 x + 5 y,\\ \frac{\partial }{\partial y}(10x^3-5x^2+5xy+5y^2+8)&=5x + 10 y. \end{align*}
Thus
$$\nabla f \left(x,y\right)=\left(30 x^{2} - 10 x + 5 y,5 x + 10 y\right)$$
Is my process correct so far? Also, to calculate $\nabla f(0,0)$, would
$$\nabla f \left(0,0\right)=\left(0,0\right)$$
as $(0,0)$ would provide no values?
The question is answered in the comment, let me just briefly summarize:
You have calculated correctly the gradient vector $\nabla f$ using one of the correct definition. Concerning $\nabla f(0,0)$, it seems to be a confusion in notation. As pointed out in the comment, $\nabla f(0,0)$ means that one first calculate $\nabla f = \nabla f(x, y)$, then evaluate at $(x, y) = (0,0)$. Since
$$ \nabla f \left(x,y\right)=\left(30 x^{2} - 10 x + 5 y,5 x + 10 y\right),$$
we have
$$ \nabla f \left(0,0\right)=\left(30 (0)^{2} - 10 (0) + 5 (0),5 (0)+ 10 (0)\right) = (0,0).$$