[tex]\dfrac{\mathrm d}{\mathrm dt}\left[2x^2+y^3\right]=\dfrac{\mathrm d}{\mathrm dt}[10]\implies 4x\dfrac{\mathrm dx}{\mathrm dt}+3y^2\dfrac{\mathrm dy}{\mathrm dt}=0[/tex]
When [tex]x=1[/tex], the original equation tells you that you have
[tex]2\times1^2+y^3=10\implies y^3=8\implies y=2[/tex]
You also know that when [tex]x=1[/tex], you have [tex]\dfrac{\mathrm dy}{\mathrm dt}=3[/tex]. Substituting everything you know into the differentiated equation, you get
[tex]4\times1\times\dfrac{\mathrm dx}{\mathrm dt}+3\times2^2\times3=0\implies\dfrac{\mathrm dx}{\mathrm dt}=-9[/tex]
So [tex]x[/tex] is changing at a rate of [tex]-9\text{ units/min}[/tex].
