For this question, we're working in $\mathbb{R}[x,y]$. To show the two sets are equivalent, I want to show that $$\langle 2x^2+3y^2-11,x^2-y^2-3 \rangle\subseteq\langle x^2-4, y^2-1 \rangle$$ and $$\langle 2x^2+3y^2-11,x^2-y^2-3 \rangle\supseteq\langle x^2-4, y^2-1 \rangle.$$ So I let $f=\alpha(2x^2+3y^2-11)+\beta(x^2-y^2-3)$ (where $\alpha,\beta$ scalars) and showed that $$f=(2\alpha+\beta)(x^2-4)+(3\alpha-\beta)(y^2-1).$$ I got this by expanding and seeing that things happened to work out.
I'm not sure how to do the converse. If I try the same thing, I get $\alpha x^2-4\alpha+\beta y^2-\beta$ and I don't know how to proceed.
You need to show that $x^2-4$ and $y^2-1$ are linear combinations of $2x^2+3y^2-11$ and $x^2-y^2-3$; to achieve this you can try and cancel the coefficients of $y^2$ and $x^2$ in $2x^2-3y^2-11$ using $x^2-y^2-3$.
More explicitly, $$2x^2+3y^2-11+3(x^2-y^2-3)=5x^2-20=5(x^2-4)$$ and $$2x^2+3y^2-11-2(x^2-y^2-3)=5y^2-5=5(y^2-1),$$ thus dividing each equations by $5$ gives you $x^2-4$ and $y^2-1$ as linear combinations of $2x+3y^2-11$ and $x^2-y^2-3$, and thus members of $\langle 2x+3y^2-11,x^2-y^2-3 \rangle$, which implies the inclusion you want.
Incidentally this is more a problem of linear algebra than polynomials : if you put $X=x^2$ and $Y=y^2$, then you want to show that $\langle 2X+3Y-11,X-Y-3 \rangle=\langle X-4, Y-1 \rangle$, which is the same as proving that the system $$\left\{ \begin{array}{}2X+3Y & = & 11\\ X-Y & = & 3\end{array}\right.$$has $(4,1)$ as a unique solution. Or more precisely, showing that the system above as this solution is equivalent to showing the equality $\langle 2X+3Y-11,X-Y-3 \rangle=\langle X-4, Y-1 \rangle$ with each member seen as a vector subspace of $\mathbb{R}[X,Y]$ generated by two elements; this implies it is also true if you consider them as ideals generated by two elements, but the ideals generated could be equal without the vector subspaces being so.