By differentiating implicitly with respect to $x$ both sides of the implicit equation
$$
\begin{equation*}
2x^{2}+y^{2}=33,\tag{1}
\end{equation*}
$$
since the derivatives of both sides should be equal we get successively:
$$
\begin{eqnarray*}
&&\frac{d}{dx}\left( 2x^{2}+y^{2}\right) =\frac{d}{dx}\left( 33\right) \\
&\Rightarrow &\frac{d}{dx}\left( 2x^{2}+y^{2}\right) =0 \\
&\Leftrightarrow &\frac{d}{dx}\left( 2x^{2}\right) +\frac{d}{dx}\left(
y^{2}\right) =0 \\
&\Leftrightarrow &4x+2y\frac{dy}{dx}=0,\qquad \frac{d}{dx}\left(
y^{2}\right) =2y\frac{dy}{dx}\text{ by the chain rule} \\
&\Leftrightarrow &\frac{dy}{dx}=-\frac{4x}{2y}=-\frac{2x}{y}\tag{2} \\
&\Rightarrow &\left. \frac{dy}{dx}\right\vert _{x=2,y=5}=-\frac{4}{5}.\tag{3}
\end{eqnarray*}
$$
The equation of the tangent line at $(2,5)$ is
$$
\begin{equation*}
y-5=-\frac{4}{5}(x-2),\tag{4}
\end{equation*}
$$
while the equation of the normal line to the curve $2x^{2}+y^{2}=33$ at $(2,5)$ is
$$
\begin{equation*}
y-5=\frac{5}{4}(x-2)\Leftrightarrow y=\frac{5}{4}x+\frac{5}{2},\tag{5}
\end{equation*}
$$
because the slope $m$ of the tangent line and the slope $m^{\prime }$ of the normal line are related by $mm^{\prime }=-1$.
ADDED. In a more general case, when we have a differentiable implicit
function $F(x,y)=0$, let $y=f(x)$ denote the function such that $F(x,f(x))\equiv 0\quad$ ($f(x)$ does not need to be explicitly known). If we differentiate both sides of $F(x,y)=0$ and apply the chain rule, we get the following total derivative with respect to $x$:
$$\frac{dF}{dx}=\frac{\partial F}{\partial x}+\frac{
\partial F}{\partial y}\frac{dy}{dx}\equiv 0.\tag{A}$$
Solving $(\mathrm{A})$ for $\frac{dy}{dx}$, gives us the following formula
$$\frac{dy}{dx}=-\frac{\partial F}{\partial x}/\frac{
\partial F}{\partial y}.\tag{B}$$
Best Answer
We need to express the derivative of the functions in terms of both $x$ and $y$ (in our answers), rather than just $x$. This is how it works:
$$a) 2x^3 = 2y^2+5$$ $$\frac{d}{dx} 2x^3 = \frac{d}{dx}2y^2 + \frac{d}{dx}5$$ $$6x = \frac{d}{dx}2y^2+0 \implies 6x = \frac{d}{dx}2y^2$$ Here, we need to use the Chain Rule to simplify $\frac{d}{dx}2y^2$. $$\frac{du}{dx} = \frac{du}{dy}\cdot\frac{dy}{dx}$$ $$\frac{d}{dx}2y^2 = \frac{d}{dy}2y^2\cdot\frac{dy}{dx}$$ $$\frac{d}{dx}2y^2 = 4y\cdot\frac{dy}{dx}$$ Now, we can continue.
$$6x = 4y\cdot\frac{dy}{dx}$$ $$\frac{dy}{dx} = \frac{6x}{4y} \implies \frac{dy}{dx} = \frac{3x}{2y}$$ We used the derivatives of both sides in order to differentiate implicitly. As you can see, it is expressed in terms of both x and y, just as the question asked. Repeating this process for the other questions will lead to the answer. $$b) 1= 3x+2x^2y^2$$ $$\frac{d}{dx}1 = \frac{d}{dx}3x+\frac{d}{dx}2x^2y^2$$ $$0 = 3+\frac{d}{dx}2x^2y^2$$ $$-3 = \frac{d}{dx}2x^2y^2$$ To find $\frac{d}{dx}2x^2y^2$, use the Product Rule. $$\frac{d}{dx}2x^2y^2 = (\frac{d}{dx}2x^2)\cdot y^2+(\frac{d}{dx}y^2)\cdot 2x^2 \implies \frac{d}{dx}2x^2y^2 = 4xy^2+2x^2\cdot 2y\cdot \frac{dy}{dx}$$ $$\frac{d}{dx}2x^2y^2 = 4xy^2+4x^2y\cdot\frac{dy}{dx}$$ Back to the question, we can quickly reach the answer. $$-3 = 4x^2y+4xy^2\cdot\frac{dy}{dx} \implies -3-4x^2y = 4xy^2\cdot\frac{dy}{dx} \implies \frac{dy}{dx} = \frac{-3-4x^2y}{4xy^2} = -\frac{3+4x^2y}{4xy^2}$$ $$c) x^2 = (4x^{2}y^{3}+1)^2$$ $$\frac{d}{dx}x^2 = \frac{d}{dx}(4x^{2}y^{3}+1)^2$$ $$2x = \frac{d}{dx}(4x^{2}y^{3}+1)^2$$ Now, $\frac{d}{dx}(4x^{2}y^{3}+1)^2$ should be calculated. Using the Chain Rule, $$\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$$ $$y = x^2 = (4x^{2}y^{3}+1)^2$$ $$u = (4x^{2}y^{3}+1)$$ $$\frac{d}{dx}(4x^{2}y^{3}+1)^2 = 2(4x^{2}y^{3}+1)\cdot\frac{d}{dx}(4x^{2}y^{3}+1)$$ $$\frac{d}{dx}(4x^{2}y^{3}+1)^2 = 2(4x^{2}y^{3}+1)\cdot (8xy^3+12x^{2}y^{2})\cdot\frac{dy}{dx}$$ Back to the question. $$2x = 2(4x^{2}y^{3}+1)\cdot (8xy^3+12x^{2}y^{2})\cdot\frac{dy}{dx}$$ $$\implies \frac{dy}{dx} = \frac{2x}{2(4x^{2}y^{3}+1)\cdot (8xy^3+12x^{2}y^{2})}$$ $$\implies \frac{dy}{dx} = \frac{x}{(4x^{2}y^{3}+1)\cdot (8xy^3+12x^{2}y^{2})}$$ For question 2, the process is repeated twice with substitution at the end. $$2) 4y^2+2 = 3x^2$$ $$\frac{d}{dx}4y^2+\frac{d}{dx}2 = \frac{d}{dx}3x^2$$ $$\frac{d}{dx}4y^2 + 0 = 6x \implies \frac{d}{dx}4y^2 = 6x$$ Using the Chain Rule, we get $$8y\cdot\frac{dy}{dx} = 6x \implies \frac{dy}{dx} = \frac{6x}{8y} = \frac{3x}{4y}$$ But the question wanted the double derivative, so the process repeats. $$\frac{dy}{dx}\frac{3x}{4y} = \frac{\frac{d}{dx}(3x)\cdot 4y -\frac{d}{dx}(4y)\cdot 3x}{(4y)^2}$$ $$\frac{d}{dx}\frac{3x}{4y} = \frac{12y-12x\cdot\frac{dy}{dx}}{16y^2}$$ Now, plug in $\frac{3x}{4y}$ for $\frac{dy}{dx}$. $$\frac{d}{dx}\frac{3x}{4y} = \frac{12y-12x\cdot\frac{3x}{4y}}{16y^2}$$ Simplify the right side and you’ll get this. $$\frac{d}{dx}\frac{3x}{4y} = \frac{12y^2-9x^2}{16y^3}$$ Therefore, this is the second derivative. $$\frac{d^2y}{dx^2} = \frac{12y^2-9x^2}{16y^3}$$