We want a function relating the current time to the size of the lily pad; let's call it $f$. From the problem, we know that
$f(60\ \mathrm{seconds})=1\ \mathrm{pond}$
$f(x\ \mathrm{seconds})=1/4\ \mathrm{pond}$
so we can indeed write an equation like the one you want by solving for $1\ \mathrm{pond}$ in each equation; then:
$\frac{f(60\ \mathrm{seconds})}1=\frac{f(x\ \mathrm{seconds})}{1/4}$
But note the mediating $f$ that you are missing in your equation! If we assumed $f(x)=x$, then we would get your equation; but the problem also tells us that the lily pad doubles in size every second, that is, that:
$f((x+1)\ \mathrm{seconds}) = 2f(x\ \mathrm{seconds})$
If we choose $f(x)=x$, then this equality is not validated, since $(x+1)\ \mathrm{seconds}=2x\ \mathrm{seconds}$ is not validated.
Luckily we can make progress even without assuming $f(x)=x$. Simplifying the corrected version of your equation, we have:
$f(60\ \mathrm{seconds})=4f(x\ \mathrm{seconds})$
Now we can apply the other equation given in the problem twice:
$f(60\ \mathrm{seconds})=4f(x\ \mathrm{seconds})$
$\phantom{f(60\ \mathrm{seconds})}=2(2f(x\ \mathrm{seconds}))$
$\phantom{f(60\ \mathrm{seconds})}=2f((x+1)\ \mathrm{seconds})$
$\phantom{f(60\ \mathrm{seconds})}=f((x+2)\ \mathrm{seconds})$
Then we can conclude that $60=x+2$ would be sufficient to validate this equation, so $x=58$ is one possible solution.